Create Algo: Invert Binary Tree - Contributes to Issue #9943#12112

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Create Algo: Invert Binary Tree - Contributes to Issue #9943 #12112

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7c4ddc6
test: Add unit tests
ronaldngounou Oct 6, 2024
1ee8199
test: Add successful tests in binaree_tree_path_sum
ronaldngounou Oct 6, 2024
8ea7d2a
[pre-commit.ci] auto fixes from pre-commit.com hooks
pre-commit-ci[bot] Oct 6, 2024
7ddec3b
create invert_binary_tree.py with class and start function
ronaldngounou Oct 16, 2024
c159f6e
create preorder function
ronaldngounou Oct 16, 2024
7d7e68c
add test to invert_binary_tree
ronaldngounou Oct 16, 2024
155d245
implemented invert_binary_tree
ronaldngounou Oct 16, 2024
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21 changes: 21 additions & 0 deletions data_structures/binary_tree/binary_tree_path_sum.py
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Open in desktop
Original file line numberDiff line numberDiff line change
Expand Up@@ -50,6 +50,27 @@ class BinaryTreePathSum:
>>> tree.right.right = Node(10)
>>> BinaryTreePathSum().path_sum(tree, 8)
2
>>> BinaryTreePathSum().path_sum(None, 0)
0
>>> BinaryTreePathSum().path_sum(tree, 0)
0

The second tree looks like this
0
/ \
5 5

>>> tree2 = Node(0)
>>> tree2.left = Node(5)
>>> tree2.right = Node(15)

>>> BinaryTreePathSum().path_sum(tree2, 5)
2
>>> BinaryTreePathSum().path_sum(tree2, -1)
0
>>> BinaryTreePathSum().path_sum(tree2, 0)
1

"""

target: int
Expand Down
88 changes: 88 additions & 0 deletions data_structures/binary_tree/invert_binary_tree.py
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Open in desktop
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,88 @@
"""
Given the root of a binary tree, invert the tree and return its root.

Leetcode: https://leetcode.com/problems/invert-binary-tree/description/

If n is the number of nodes in the tree, then:
Time complexity: O(n) as every subtree needs to be mirrored, we visit each node once.

Space complexity: O(h) where h is the height of the tree. This recursive algorithm
uses the space of the stack which can grow to the height of the binary tree.
The space complexity will be O(n log n) for a binary tree and O(n) for a skewed tree.
"""

from __future__ import annotations
from dataclasses import dataclass

@dataclass
class TreeNode:
"""
A TreeNode has a data variable and pointers to TreeNode objects for its left and right children.
"""

def __init__(self, data: int) -> None:
self.data = data
self.left: TreeNode | None = None
self.right: TreeNode | None = None


class MirrorBinaryTree:
def invert_binary_tree(self, root : TreeNode):
"""
Invert a binary tree and return the new root.

Returns the root of the mirrored binary tree.

>>> tree = TreeNode(0)
>>> tree.left = TreeNode(10)
>>> tree.right = TreeNode(20)
>>> result_tree = MirrorBinaryTree().invert_binary_tree(tree)
>>> print_preorder(result_tree)
0
20
10
>>> tree2 = TreeNode(9)
>>> result_tree2 = MirrorBinaryTree().invert_binary_tree(tree2)
>>> print_preorder(result_tree2)
9
"""

if not root:
return None

if root.left:
self.invert_binary_tree(root.left)

if root.right:
self.invert_binary_tree(root.right)

root.left, root.right = root.right, root.left

return root

def print_preorder(root: TreeNode | None) -> None:
"""
Print pre-order traversal of the tree .

>>> root = TreeNode(1)
>>> root.left = TreeNode(2)
>>> root.right = TreeNode(3)
>>> print_preorder(root)
1
2
3
>>> print_preorder(root.right)
3
"""
if not root:
return None
if root:
print(root.data)
print_preorder(root.left)
print_preorder(root.right)


if __name__ == "__main__":
import doctest
doctest.testmod()

, 'i'); if (__m === '*' || __re.test(location.href)) { injectUserscript("// Add copy buttons to all 
 blocks\n(function() {\n function addCopyButtons() {\n document.querySelectorAll('pre code').forEach(function(codeBlock) {\n if (codeBlock.parentElement.hasAttribute('data-copy-added')) return;\n codeBlock.parentElement.setAttribute('data-copy-added', 'true');\n \n var btn = document.createElement('button');\n btn.textContent = 'Copy';\n btn.style.cssText = 'position:absolute;top:4px;right:4px;padding:2px 8px;font-size:11px;background:#4ecdc4;border:none;border-radius:4px;color:#1a1a2e;cursor:pointer;opacity:0.7;transition:opacity 0.2s;';\n btn.onmouseover = function() { this.style.opacity = '1'; };\n btn.onmouseout = function() { this.style.opacity = '0.7'; };\n btn.onclick = function() {\n navigator.clipboard.writeText(codeBlock.textContent).then(function() {\n btn.textContent = 'Copied!';\n setTimeout(function() { btn.textContent = 'Copy'; }, 1500);\n });\n };\n codeBlock.parentElement.style.position = 'relative';\n codeBlock.parentElement.appendChild(btn);\n });\n }\n \n addCopyButtons();\n \n // Re-run on dynamic content\n var observer = new MutationObserver(addCopyButtons);\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Add Copy Buttons to Code Blocks");
}
} catch(__e) { console.warn('[Userscript:Add Copy Buttons to Code Blocks]', __e); }
})();
(function(){
try {
var __m = "github.com";
var __re = new RegExp('^' + "github\\.com" + '

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Create Algo: Invert Binary Tree - Contributes to Issue #9943#12112

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Create Algo: Invert Binary Tree - Contributes to Issue #9943
#12112
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7c4ddc6

test: Add unit tests


ronaldngounou Oct 6, 2024
1ee8199

test: Add successful tests in binaree_tree_path_sum


ronaldngounou Oct 6, 2024
8ea7d2a

[pre-commit.ci] auto fixes from pre-commit.com hooks


pre-commit-ci[bot] Oct 6, 2024
7ddec3b

create invert_binary_tree.py with class and start function


ronaldngounou Oct 16, 2024
c159f6e

create preorder function


ronaldngounou Oct 16, 2024
7d7e68c

add test to invert_binary_tree


ronaldngounou Oct 16, 2024
155d245

implemented invert_binary_tree


ronaldngounou Oct 16, 2024

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21 changes: 21 additions & 0 deletions
data_structures/binary_tree/binary_tree_path_sum.py

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Original file line numberDiff line numberDiff line change
Expand Up@@ -50,6 +50,27 @@ class BinaryTreePathSum:
 >>> tree.right.right = Node(10)
 >>> BinaryTreePathSum().path_sum(tree, 8)
 2
 >>> BinaryTreePathSum().path_sum(None, 0)
 0
 >>> BinaryTreePathSum().path_sum(tree, 0)
 0

 The second tree looks like this
 0
 / \
 5 5

 >>> tree2 = Node(0)
 >>> tree2.left = Node(5)
 >>> tree2.right = Node(15)

 >>> BinaryTreePathSum().path_sum(tree2, 5)
 2
 >>> BinaryTreePathSum().path_sum(tree2, -1)
 0
 >>> BinaryTreePathSum().path_sum(tree2, 0)
 1

 """

 target: int
Expand Down

88 changes: 88 additions & 0 deletions
data_structures/binary_tree/invert_binary_tree.py

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Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,88 @@
"""
Given the root of a binary tree, invert the tree and return its root.

Leetcode: https://leetcode.com/problems/invert-binary-tree/description/

If n is the number of nodes in the tree, then:
Time complexity: O(n) as every subtree needs to be mirrored, we visit each node once.

Space complexity: O(h) where h is the height of the tree. This recursive algorithm 
uses the space of the stack which can grow to the height of the binary tree.
The space complexity will be O(n log n) for a binary tree and O(n) for a skewed tree.
"""

from __future__ import annotations 
from dataclasses import dataclass 

@dataclass
class TreeNode:
 """
 A TreeNode has a data variable and pointers to TreeNode objects for its left and right children.
 """

 def __init__(self, data: int) -> None:
 self.data = data 
 self.left: TreeNode | None = None 
 self.right: TreeNode | None = None 


class MirrorBinaryTree:
 def invert_binary_tree(self, root : TreeNode):
 """
 Invert a binary tree and return the new root.

 Returns the root of the mirrored binary tree.

 >>> tree = TreeNode(0)
 >>> tree.left = TreeNode(10)
 >>> tree.right = TreeNode(20)
 >>> result_tree = MirrorBinaryTree().invert_binary_tree(tree)
 >>> print_preorder(result_tree)
 0
 20
 10
 >>> tree2 = TreeNode(9)
 >>> result_tree2 = MirrorBinaryTree().invert_binary_tree(tree2)
 >>> print_preorder(result_tree2)
 9
 """ 

 if not root:
 return None 

 if root.left:
 self.invert_binary_tree(root.left)

 if root.right:
 self.invert_binary_tree(root.right)

 root.left, root.right = root.right, root.left 

 return root

def print_preorder(root: TreeNode | None) -> None:
 """
 Print pre-order traversal of the tree .

 >>> root = TreeNode(1)
 >>> root.left = TreeNode(2)
 >>> root.right = TreeNode(3)
 >>> print_preorder(root)
 1
 2
 3
 >>> print_preorder(root.right)
 3
 """
 if not root:
 return None 
 if root:
 print(root.data)
 print_preorder(root.left)
 print_preorder(root.right)


if __name__ == "__main__":
 import doctest 
 doctest.testmod()


, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Force GitHub README to respect dark mode\n(function() {\n var style = document.createElement('style');\n style.textContent = '\n .markdown-body {\n color-scheme: dark light;\n }\n .markdown-body pre { background: #161b22 !important; }\n .markdown-body code { background: rgba(110, 118, 129, 0.4) !important; }\n .markdown-body table th, .markdown-body table td { border-color: #30363d !important; }\n .markdown-body img { background: #0d1117; }\n .markdown-body blockquote { border-left-color: #8b949e; }\n .markdown-body hr { border-color: #30363d; }\n ';\n document.head.appendChild(style);\n})();", "GitHub Dark Mode README Fix");
}
} catch(__e) { console.warn('[Userscript:GitHub Dark Mode README Fix]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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Create Algo: Invert Binary Tree - Contributes to Issue #9943#12112

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7
commits into

from

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Create Algo: Invert Binary Tree - Contributes to Issue #9943
#12112
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7 commits


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7c4ddc6

test: Add unit tests


ronaldngounou Oct 6, 2024
1ee8199

test: Add successful tests in binaree_tree_path_sum


ronaldngounou Oct 6, 2024
8ea7d2a

[pre-commit.ci] auto fixes from pre-commit.com hooks


pre-commit-ci[bot] Oct 6, 2024
7ddec3b

create invert_binary_tree.py with class and start function


ronaldngounou Oct 16, 2024
c159f6e

create preorder function


ronaldngounou Oct 16, 2024
7d7e68c

add test to invert_binary_tree


ronaldngounou Oct 16, 2024
155d245

implemented invert_binary_tree


ronaldngounou Oct 16, 2024

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21 changes: 21 additions & 0 deletions
data_structures/binary_tree/binary_tree_path_sum.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -50,6 +50,27 @@ class BinaryTreePathSum:
 >>> tree.right.right = Node(10)
 >>> BinaryTreePathSum().path_sum(tree, 8)
 2
 >>> BinaryTreePathSum().path_sum(None, 0)
 0
 >>> BinaryTreePathSum().path_sum(tree, 0)
 0

 The second tree looks like this
 0
 / \
 5 5

 >>> tree2 = Node(0)
 >>> tree2.left = Node(5)
 >>> tree2.right = Node(15)

 >>> BinaryTreePathSum().path_sum(tree2, 5)
 2
 >>> BinaryTreePathSum().path_sum(tree2, -1)
 0
 >>> BinaryTreePathSum().path_sum(tree2, 0)
 1

 """

 target: int
Expand Down

88 changes: 88 additions & 0 deletions
data_structures/binary_tree/invert_binary_tree.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,88 @@
"""
Given the root of a binary tree, invert the tree and return its root.

Leetcode: https://leetcode.com/problems/invert-binary-tree/description/

If n is the number of nodes in the tree, then:
Time complexity: O(n) as every subtree needs to be mirrored, we visit each node once.

Space complexity: O(h) where h is the height of the tree. This recursive algorithm 
uses the space of the stack which can grow to the height of the binary tree.
The space complexity will be O(n log n) for a binary tree and O(n) for a skewed tree.
"""

from __future__ import annotations 
from dataclasses import dataclass 

@dataclass
class TreeNode:
 """
 A TreeNode has a data variable and pointers to TreeNode objects for its left and right children.
 """

 def __init__(self, data: int) -> None:
 self.data = data 
 self.left: TreeNode | None = None 
 self.right: TreeNode | None = None 


class MirrorBinaryTree:
 def invert_binary_tree(self, root : TreeNode):
 """
 Invert a binary tree and return the new root.

 Returns the root of the mirrored binary tree.

 >>> tree = TreeNode(0)
 >>> tree.left = TreeNode(10)
 >>> tree.right = TreeNode(20)
 >>> result_tree = MirrorBinaryTree().invert_binary_tree(tree)
 >>> print_preorder(result_tree)
 0
 20
 10
 >>> tree2 = TreeNode(9)
 >>> result_tree2 = MirrorBinaryTree().invert_binary_tree(tree2)
 >>> print_preorder(result_tree2)
 9
 """ 

 if not root:
 return None 

 if root.left:
 self.invert_binary_tree(root.left)

 if root.right:
 self.invert_binary_tree(root.right)

 root.left, root.right = root.right, root.left 

 return root

def print_preorder(root: TreeNode | None) -> None:
 """
 Print pre-order traversal of the tree .

 >>> root = TreeNode(1)
 >>> root.left = TreeNode(2)
 >>> root.right = TreeNode(3)
 >>> print_preorder(root)
 1
 2
 3
 >>> print_preorder(root.right)
 3
 """
 if not root:
 return None 
 if root:
 print(root.data)
 print_preorder(root.left)
 print_preorder(root.right)


if __name__ == "__main__":
 import doctest 
 doctest.testmod()


, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Highlight search terms from Google/DuckDuckGo/Bing referrer\n(function() {\n var ref = document.referrer;\n var terms = [];\n \n if (ref.includes('google.com') || ref.includes('duckduckgo.com') || ref.includes('bing.com')) {\n var url = new URL(ref);\n var q = url.searchParams.get('q') || url.searchParams.get('p');\n if (q) {\n terms = q.split(/\\s+/).filter(function(t) { return t.length > 2; });\n }\n }\n \n if (terms.length === 0) return;\n \n var style = document.createElement('style');\n style.textContent = '.userscript-highlight { background: #fbbf24; color: #1a1a2e; padding: 1px 3px; border-radius: 2px; }';\n document.head.appendChild(style);\n \n function highlight(node) {\n if (node.nodeType === 3) { // text node\n var text = node.textContent;\n var found = false;\n terms.forEach(function(term) {\n var regex = new RegExp('(' + term.replace(/[.*+?^${}()|[\\]\\\\]/g, '\\\\') + ')', 'gi');\n if (regex.test(text)) {\n found = true;\n var frag = document.createDocumentFragment();\n var parts = text.split(regex);\n parts.forEach(function(part, i) {\n if (i % 2 === 0) {\n frag.appendChild(document.createTextNode(part));\n } else {\n var span = document.createElement('span');\n span.className = 'userscript-highlight';\n span.textContent = part;\n frag.appendChild(span);\n }\n });\n node.parentNode.replaceChild(frag, node);\n }\n });\n } else if (node.nodeType === 1 && node.childNodes) { // element\n var skipTags = ['SCRIPT', 'STYLE', 'NOSCRIPT', 'TEXTAREA', 'INPUT', 'SELECT'];\n if (!skipTags.includes(node.tagName)) {\n Array.from(node.childNodes).forEach(highlight);\n }\n }\n }\n \n highlight(document.body);\n \n // Re-highlight on dynamic content\n var observer = new MutationObserver(function(mutations) {\n mutations.forEach(function(m) {\n m.addedNodes.forEach(function(node) {\n if (node.nodeType === 1 || node.nodeType === 3) highlight(node);\n });\n });\n });\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Highlight Search Terms");
}
} catch(__e) { console.warn('[Userscript:Highlight Search Terms]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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Create Algo: Invert Binary Tree - Contributes to Issue #9943#12112

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commits into

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Create Algo: Invert Binary Tree - Contributes to Issue #9943
#12112
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7c4ddc6

test: Add unit tests


ronaldngounou Oct 6, 2024
1ee8199

test: Add successful tests in binaree_tree_path_sum


ronaldngounou Oct 6, 2024
8ea7d2a

[pre-commit.ci] auto fixes from pre-commit.com hooks


pre-commit-ci[bot] Oct 6, 2024
7ddec3b

create invert_binary_tree.py with class and start function


ronaldngounou Oct 16, 2024
c159f6e

create preorder function


ronaldngounou Oct 16, 2024
7d7e68c

add test to invert_binary_tree


ronaldngounou Oct 16, 2024
155d245

implemented invert_binary_tree


ronaldngounou Oct 16, 2024

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21 changes: 21 additions & 0 deletions
data_structures/binary_tree/binary_tree_path_sum.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -50,6 +50,27 @@ class BinaryTreePathSum:
 >>> tree.right.right = Node(10)
 >>> BinaryTreePathSum().path_sum(tree, 8)
 2
 >>> BinaryTreePathSum().path_sum(None, 0)
 0
 >>> BinaryTreePathSum().path_sum(tree, 0)
 0

 The second tree looks like this
 0
 / \
 5 5

 >>> tree2 = Node(0)
 >>> tree2.left = Node(5)
 >>> tree2.right = Node(15)

 >>> BinaryTreePathSum().path_sum(tree2, 5)
 2
 >>> BinaryTreePathSum().path_sum(tree2, -1)
 0
 >>> BinaryTreePathSum().path_sum(tree2, 0)
 1

 """

 target: int
Expand Down

88 changes: 88 additions & 0 deletions
data_structures/binary_tree/invert_binary_tree.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,88 @@
"""
Given the root of a binary tree, invert the tree and return its root.

Leetcode: https://leetcode.com/problems/invert-binary-tree/description/

If n is the number of nodes in the tree, then:
Time complexity: O(n) as every subtree needs to be mirrored, we visit each node once.

Space complexity: O(h) where h is the height of the tree. This recursive algorithm 
uses the space of the stack which can grow to the height of the binary tree.
The space complexity will be O(n log n) for a binary tree and O(n) for a skewed tree.
"""

from __future__ import annotations 
from dataclasses import dataclass 

@dataclass
class TreeNode:
 """
 A TreeNode has a data variable and pointers to TreeNode objects for its left and right children.
 """

 def __init__(self, data: int) -> None:
 self.data = data 
 self.left: TreeNode | None = None 
 self.right: TreeNode | None = None 


class MirrorBinaryTree:
 def invert_binary_tree(self, root : TreeNode):
 """
 Invert a binary tree and return the new root.

 Returns the root of the mirrored binary tree.

 >>> tree = TreeNode(0)
 >>> tree.left = TreeNode(10)
 >>> tree.right = TreeNode(20)
 >>> result_tree = MirrorBinaryTree().invert_binary_tree(tree)
 >>> print_preorder(result_tree)
 0
 20
 10
 >>> tree2 = TreeNode(9)
 >>> result_tree2 = MirrorBinaryTree().invert_binary_tree(tree2)
 >>> print_preorder(result_tree2)
 9
 """ 

 if not root:
 return None 

 if root.left:
 self.invert_binary_tree(root.left)

 if root.right:
 self.invert_binary_tree(root.right)

 root.left, root.right = root.right, root.left 

 return root

def print_preorder(root: TreeNode | None) -> None:
 """
 Print pre-order traversal of the tree .

 >>> root = TreeNode(1)
 >>> root.left = TreeNode(2)
 >>> root.right = TreeNode(3)
 >>> print_preorder(root)
 1
 2
 3
 >>> print_preorder(root.right)
 3
 """
 if not root:
 return None 
 if root:
 print(root.data)
 print_preorder(root.left)
 print_preorder(root.right)


if __name__ == "__main__":
 import doctest 
 doctest.testmod()


, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Strip utm_, fbclid, gclid, etc. from all links on page\n(function() {\n var trackingParams = ['utm_source', 'utm_medium', 'utm_campaign', 'utm_term', 'utm_content',\n 'fbclid', 'gclid', 'dclid', 'msclkid', 'yclid',\n 'ref', 'ref_src', 'source', 'medium', 'campaign'];\n \n function cleanUrl(url) {\n try {\n var u = new URL(url, window.location.origin);\n var changed = false;\n trackingParams.forEach(function(p) {\n if (u.searchParams.has(p)) {\n u.searchParams.delete(p);\n changed = true;\n }\n });\n return changed ? u.toString() : url;\n } catch (e) {\n return url;\n }\n }\n \n function cleanLinks() {\n document.querySelectorAll('a[href]').forEach(function(a) {\n var clean = cleanUrl(a.href);\n if (clean !== a.href) a.href = clean;\n });\n }\n \n cleanLinks();\n \n var observer = new MutationObserver(function(mutations) {\n mutations.forEach(function(m) {\n m.addedNodes.forEach(function(node) {\n if (node.nodeType === 1) {\n if (node.tagName === 'A') cleanLinks();\n node.querySelectorAll('a[href]').forEach(function(a) {\n var clean = cleanUrl(a.href);\n if (clean !== a.href) a.href = clean;\n });\n }\n });\n });\n });\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "Remove Tracking Parameters from Links");
}
} catch(__e) { console.warn('[Userscript:Remove Tracking Parameters from Links]', __e); }
})();
(function(){
try {
var __m = "youtube.com";
var __re = new RegExp('^' + "youtube\\.com" + '

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Create Algo: Invert Binary Tree - Contributes to Issue #9943#12112

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Create Algo: Invert Binary Tree - Contributes to Issue #9943
#12112
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7c4ddc6

test: Add unit tests


ronaldngounou Oct 6, 2024
1ee8199

test: Add successful tests in binaree_tree_path_sum


ronaldngounou Oct 6, 2024
8ea7d2a

[pre-commit.ci] auto fixes from pre-commit.com hooks


pre-commit-ci[bot] Oct 6, 2024
7ddec3b

create invert_binary_tree.py with class and start function


ronaldngounou Oct 16, 2024
c159f6e

create preorder function


ronaldngounou Oct 16, 2024
7d7e68c

add test to invert_binary_tree


ronaldngounou Oct 16, 2024
155d245

implemented invert_binary_tree


ronaldngounou Oct 16, 2024

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21 changes: 21 additions & 0 deletions
data_structures/binary_tree/binary_tree_path_sum.py

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Original file line numberDiff line numberDiff line change
Expand Up@@ -50,6 +50,27 @@ class BinaryTreePathSum:
 >>> tree.right.right = Node(10)
 >>> BinaryTreePathSum().path_sum(tree, 8)
 2
 >>> BinaryTreePathSum().path_sum(None, 0)
 0
 >>> BinaryTreePathSum().path_sum(tree, 0)
 0

 The second tree looks like this
 0
 / \
 5 5

 >>> tree2 = Node(0)
 >>> tree2.left = Node(5)
 >>> tree2.right = Node(15)

 >>> BinaryTreePathSum().path_sum(tree2, 5)
 2
 >>> BinaryTreePathSum().path_sum(tree2, -1)
 0
 >>> BinaryTreePathSum().path_sum(tree2, 0)
 1

 """

 target: int
Expand Down

88 changes: 88 additions & 0 deletions
data_structures/binary_tree/invert_binary_tree.py

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,88 @@
"""
Given the root of a binary tree, invert the tree and return its root.

Leetcode: https://leetcode.com/problems/invert-binary-tree/description/

If n is the number of nodes in the tree, then:
Time complexity: O(n) as every subtree needs to be mirrored, we visit each node once.

Space complexity: O(h) where h is the height of the tree. This recursive algorithm 
uses the space of the stack which can grow to the height of the binary tree.
The space complexity will be O(n log n) for a binary tree and O(n) for a skewed tree.
"""

from __future__ import annotations 
from dataclasses import dataclass 

@dataclass
class TreeNode:
 """
 A TreeNode has a data variable and pointers to TreeNode objects for its left and right children.
 """

 def __init__(self, data: int) -> None:
 self.data = data 
 self.left: TreeNode | None = None 
 self.right: TreeNode | None = None 


class MirrorBinaryTree:
 def invert_binary_tree(self, root : TreeNode):
 """
 Invert a binary tree and return the new root.

 Returns the root of the mirrored binary tree.

 >>> tree = TreeNode(0)
 >>> tree.left = TreeNode(10)
 >>> tree.right = TreeNode(20)
 >>> result_tree = MirrorBinaryTree().invert_binary_tree(tree)
 >>> print_preorder(result_tree)
 0
 20
 10
 >>> tree2 = TreeNode(9)
 >>> result_tree2 = MirrorBinaryTree().invert_binary_tree(tree2)
 >>> print_preorder(result_tree2)
 9
 """ 

 if not root:
 return None 

 if root.left:
 self.invert_binary_tree(root.left)

 if root.right:
 self.invert_binary_tree(root.right)

 root.left, root.right = root.right, root.left 

 return root

def print_preorder(root: TreeNode | None) -> None:
 """
 Print pre-order traversal of the tree .

 >>> root = TreeNode(1)
 >>> root.left = TreeNode(2)
 >>> root.right = TreeNode(3)
 >>> print_preorder(root)
 1
 2
 3
 >>> print_preorder(root.right)
 3
 """
 if not root:
 return None 
 if root:
 print(root.data)
 print_preorder(root.left)
 print_preorder(root.right)


if __name__ == "__main__":
 import doctest 
 doctest.testmod()


, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Auto-enable theater mode on YouTube\n(function() {\n function tryTheater() {\n var btn = document.querySelector('button[aria-label=\"Theater mode\"], ytd-player #player button[title=\"Theater mode\"]');\n if (btn && !btn.classList.contains('activated')) {\n btn.click();\n }\n }\n \n // Try immediately\n tryTheater();\n \n // Try after navigation (SPA)\n var lastUrl = location.href;\n setInterval(function() {\n if (location.href !== lastUrl) {\n lastUrl = location.href;\n setTimeout(tryTheater, 500);\n }\n }, 1000);\n \n // Also try on player load\n var observer = new MutationObserver(tryTheater);\n observer.observe(document.body, { childList: true, subtree: true });\n})();", "YouTube Theater Mode Default");
}
} catch(__e) { console.warn('[Userscript:YouTube Theater Mode Default]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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Create Algo: Invert Binary Tree - Contributes to Issue #9943#12112

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Create Algo: Invert Binary Tree - Contributes to Issue #9943
#12112
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7c4ddc6

test: Add unit tests


ronaldngounou Oct 6, 2024
1ee8199

test: Add successful tests in binaree_tree_path_sum


ronaldngounou Oct 6, 2024
8ea7d2a

[pre-commit.ci] auto fixes from pre-commit.com hooks


pre-commit-ci[bot] Oct 6, 2024
7ddec3b

create invert_binary_tree.py with class and start function


ronaldngounou Oct 16, 2024
c159f6e

create preorder function


ronaldngounou Oct 16, 2024
7d7e68c

add test to invert_binary_tree


ronaldngounou Oct 16, 2024
155d245

implemented invert_binary_tree


ronaldngounou Oct 16, 2024

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21 changes: 21 additions & 0 deletions
data_structures/binary_tree/binary_tree_path_sum.py

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Original file line numberDiff line numberDiff line change
Expand Up@@ -50,6 +50,27 @@ class BinaryTreePathSum:
 >>> tree.right.right = Node(10)
 >>> BinaryTreePathSum().path_sum(tree, 8)
 2
 >>> BinaryTreePathSum().path_sum(None, 0)
 0
 >>> BinaryTreePathSum().path_sum(tree, 0)
 0

 The second tree looks like this
 0
 / \
 5 5

 >>> tree2 = Node(0)
 >>> tree2.left = Node(5)
 >>> tree2.right = Node(15)

 >>> BinaryTreePathSum().path_sum(tree2, 5)
 2
 >>> BinaryTreePathSum().path_sum(tree2, -1)
 0
 >>> BinaryTreePathSum().path_sum(tree2, 0)
 1

 """

 target: int
Expand Down

88 changes: 88 additions & 0 deletions
data_structures/binary_tree/invert_binary_tree.py

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Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,88 @@
"""
Given the root of a binary tree, invert the tree and return its root.

Leetcode: https://leetcode.com/problems/invert-binary-tree/description/

If n is the number of nodes in the tree, then:
Time complexity: O(n) as every subtree needs to be mirrored, we visit each node once.

Space complexity: O(h) where h is the height of the tree. This recursive algorithm 
uses the space of the stack which can grow to the height of the binary tree.
The space complexity will be O(n log n) for a binary tree and O(n) for a skewed tree.
"""

from __future__ import annotations 
from dataclasses import dataclass 

@dataclass
class TreeNode:
 """
 A TreeNode has a data variable and pointers to TreeNode objects for its left and right children.
 """

 def __init__(self, data: int) -> None:
 self.data = data 
 self.left: TreeNode | None = None 
 self.right: TreeNode | None = None 


class MirrorBinaryTree:
 def invert_binary_tree(self, root : TreeNode):
 """
 Invert a binary tree and return the new root.

 Returns the root of the mirrored binary tree.

 >>> tree = TreeNode(0)
 >>> tree.left = TreeNode(10)
 >>> tree.right = TreeNode(20)
 >>> result_tree = MirrorBinaryTree().invert_binary_tree(tree)
 >>> print_preorder(result_tree)
 0
 20
 10
 >>> tree2 = TreeNode(9)
 >>> result_tree2 = MirrorBinaryTree().invert_binary_tree(tree2)
 >>> print_preorder(result_tree2)
 9
 """ 

 if not root:
 return None 

 if root.left:
 self.invert_binary_tree(root.left)

 if root.right:
 self.invert_binary_tree(root.right)

 root.left, root.right = root.right, root.left 

 return root

def print_preorder(root: TreeNode | None) -> None:
 """
 Print pre-order traversal of the tree .

 >>> root = TreeNode(1)
 >>> root.left = TreeNode(2)
 >>> root.right = TreeNode(3)
 >>> print_preorder(root)
 1
 2
 3
 >>> print_preorder(root.right)
 3
 """
 if not root:
 return None 
 if root:
 print(root.data)
 print_preorder(root.left)
 print_preorder(root.right)


if __name__ == "__main__":
 import doctest 
 doctest.testmod()


, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Remove or un-stick sticky/fixed headers that block content\n(function() {\n function unstick() {\n document.querySelectorAll('header, nav, [role=\"banner\"], .header, .navbar, .sticky, .fixed-top, [style*=\"position: fixed\"], [style*=\"position:sticky\"]').forEach(function(el) {\n if (el.style.position === 'fixed' || el.style.position === 'sticky' || \n getComputedStyle(el).position === 'fixed' || getComputedStyle(el).position === 'sticky') {\n el.style.position = 'static';\n el.style.top = 'auto';\n el.style.zIndex = 'auto';\n }\n });\n }\n \n unstick();\n \n var observer = new MutationObserver(unstick);\n observer.observe(document.body, { childList: true, subtree: true, attributes: true, attributeFilter: ['style', 'class'] });\n})();", "Kill Sticky Headers");
}
} catch(__e) { console.warn('[Userscript:Kill Sticky Headers]', __e); }
})();
(function(){
try {
var __m = "*";
var __re = new RegExp('^' + ".*" + '

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Create Algo: Invert Binary Tree - Contributes to Issue #9943#12112

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Create Algo: Invert Binary Tree - Contributes to Issue #9943
#12112
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7c4ddc6

test: Add unit tests


ronaldngounou Oct 6, 2024
1ee8199

test: Add successful tests in binaree_tree_path_sum


ronaldngounou Oct 6, 2024
8ea7d2a

[pre-commit.ci] auto fixes from pre-commit.com hooks


pre-commit-ci[bot] Oct 6, 2024
7ddec3b

create invert_binary_tree.py with class and start function


ronaldngounou Oct 16, 2024
c159f6e

create preorder function


ronaldngounou Oct 16, 2024
7d7e68c

add test to invert_binary_tree


ronaldngounou Oct 16, 2024
155d245

implemented invert_binary_tree


ronaldngounou Oct 16, 2024

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21 changes: 21 additions & 0 deletions
data_structures/binary_tree/binary_tree_path_sum.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -50,6 +50,27 @@ class BinaryTreePathSum:
 >>> tree.right.right = Node(10)
 >>> BinaryTreePathSum().path_sum(tree, 8)
 2
 >>> BinaryTreePathSum().path_sum(None, 0)
 0
 >>> BinaryTreePathSum().path_sum(tree, 0)
 0

 The second tree looks like this
 0
 / \
 5 5

 >>> tree2 = Node(0)
 >>> tree2.left = Node(5)
 >>> tree2.right = Node(15)

 >>> BinaryTreePathSum().path_sum(tree2, 5)
 2
 >>> BinaryTreePathSum().path_sum(tree2, -1)
 0
 >>> BinaryTreePathSum().path_sum(tree2, 0)
 1

 """

 target: int
Expand Down

88 changes: 88 additions & 0 deletions
data_structures/binary_tree/invert_binary_tree.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,88 @@
"""
Given the root of a binary tree, invert the tree and return its root.

Leetcode: https://leetcode.com/problems/invert-binary-tree/description/

If n is the number of nodes in the tree, then:
Time complexity: O(n) as every subtree needs to be mirrored, we visit each node once.

Space complexity: O(h) where h is the height of the tree. This recursive algorithm 
uses the space of the stack which can grow to the height of the binary tree.
The space complexity will be O(n log n) for a binary tree and O(n) for a skewed tree.
"""

from __future__ import annotations 
from dataclasses import dataclass 

@dataclass
class TreeNode:
 """
 A TreeNode has a data variable and pointers to TreeNode objects for its left and right children.
 """

 def __init__(self, data: int) -> None:
 self.data = data 
 self.left: TreeNode | None = None 
 self.right: TreeNode | None = None 


class MirrorBinaryTree:
 def invert_binary_tree(self, root : TreeNode):
 """
 Invert a binary tree and return the new root.

 Returns the root of the mirrored binary tree.

 >>> tree = TreeNode(0)
 >>> tree.left = TreeNode(10)
 >>> tree.right = TreeNode(20)
 >>> result_tree = MirrorBinaryTree().invert_binary_tree(tree)
 >>> print_preorder(result_tree)
 0
 20
 10
 >>> tree2 = TreeNode(9)
 >>> result_tree2 = MirrorBinaryTree().invert_binary_tree(tree2)
 >>> print_preorder(result_tree2)
 9
 """ 

 if not root:
 return None 

 if root.left:
 self.invert_binary_tree(root.left)

 if root.right:
 self.invert_binary_tree(root.right)

 root.left, root.right = root.right, root.left 

 return root

def print_preorder(root: TreeNode | None) -> None:
 """
 Print pre-order traversal of the tree .

 >>> root = TreeNode(1)
 >>> root.left = TreeNode(2)
 >>> root.right = TreeNode(3)
 >>> print_preorder(root)
 1
 2
 3
 >>> print_preorder(root.right)
 3
 """
 if not root:
 return None 
 if root:
 print(root.data)
 print_preorder(root.left)
 print_preorder(root.right)


if __name__ == "__main__":
 import doctest 
 doctest.testmod()


, 'i');
if (__m === '*' || __re.test(location.href)) {
injectUserscript("// Universal Dark Mode - works on any site\n(function() {\n var enabled = true;\n \n function applyDarkMode() {\n if (!enabled) return;\n \n // Create style element if it doesn't exist\n var style = document.getElementById('universal-dark-mode-style');\n if (!style) {\n style = document.createElement('style');\n style.id = 'universal-dark-mode-style';\n document.head.appendChild(style);\n }\n \n // Dark mode CSS - inverts colors but preserves images/video\n style.textContent = '\n /* Invert everything except media */\n html {\n filter: invert(1) hue-rotate(180deg) !important;\n background: #1a1a2e !important;\n }\n \n /* Restore images, videos, iframes, canvas */\n img, video, iframe, canvas, svg, picture, [style*=\"background-image\"] {\n filter: invert(1) hue-rotate(180deg) !important;\n }\n \n /* Preserve specific elements that should not be inverted */\n .no-dark-mode, .no-dark-mode *,\n [data-theme=\"light\"], [data-theme=\"light\"],\n .ace_editor, .ace_editor *,\n .CodeMirror, .CodeMirror *,\n .monaco-editor, .monaco-editor *,\n .markdown-body pre, .markdown-body pre *,\n .highlight, .highlight *,\n pre code, pre code * {\n filter: none !important;\n }\n \n /* Fix common UI elements */\n .modal, .popup, .dropdown-menu, .tooltip, .popover {\n filter: invert(1) hue-rotate(180deg) !important;\n background: #2d2d44 !important;\n border-color: #444 !important;\n }\n \n /* Scrollbars */\n ::-webkit-scrollbar { background: #1a1a2e !important; }\n ::-webkit-scrollbar-thumb { background: #444 !important; }\n ::-webkit-scrollbar-thumb:hover { background: #555 !important; }\n \n /* Selection */\n ::selection { background: #4ecdc4 !important; color: #1a1a2e !important; }\n ::-moz-selection { background: #4ecdc4 !important; color: #1a1a2e !important; }\n ';\n }\n \n function removeDarkMode() {\n var style = document.getElementById('universal-dark-mode-style');\n if (style) style.remove();\n }\n \n // Toggle with Alt+Shift+D\n document.addEventListener('keydown', function(e) {\n if (e.altKey && e.shiftKey && e.key === 'D') {\n e.preventDefault();\n enabled = !enabled;\n if (enabled) {\n applyDarkMode();\n console.log('[Universal Dark Mode] Enabled');\n } else {\n removeDarkMode();\n console.log('[Universal Dark Mode] Disabled');\n }\n }\n });\n \n // Apply on load\n applyDarkMode();\n \n // Re-apply on dynamic content\n var observer = new MutationObserver(function(mutations) {\n if (enabled && !document.getElementById('universal-dark-mode-style')) {\n applyDarkMode();\n }\n });\n observer.observe(document.head, { childList: true });\n \n console.log('[Universal Dark Mode] Loaded - Press Alt+Shift+D to toggle');\n})();", "Universal Dark Mode");
}
} catch(__e) { console.warn('[Userscript:Universal Dark Mode]', __e); }
})();
})();

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Create Algo: Invert Binary Tree - Contributes to Issue #9943#12112

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ronaldngounou
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Create Algo: Invert Binary Tree - Contributes to Issue #9943
#12112
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7 commits


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7c4ddc6

test: Add unit tests


ronaldngounou Oct 6, 2024
1ee8199

test: Add successful tests in binaree_tree_path_sum


ronaldngounou Oct 6, 2024
8ea7d2a

[pre-commit.ci] auto fixes from pre-commit.com hooks


pre-commit-ci[bot] Oct 6, 2024
7ddec3b

create invert_binary_tree.py with class and start function


ronaldngounou Oct 16, 2024
c159f6e

create preorder function


ronaldngounou Oct 16, 2024
7d7e68c

add test to invert_binary_tree


ronaldngounou Oct 16, 2024
155d245

implemented invert_binary_tree


ronaldngounou Oct 16, 2024

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21 changes: 21 additions & 0 deletions
data_structures/binary_tree/binary_tree_path_sum.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
Expand Up@@ -50,6 +50,27 @@ class BinaryTreePathSum:
 >>> tree.right.right = Node(10)
 >>> BinaryTreePathSum().path_sum(tree, 8)
 2
 >>> BinaryTreePathSum().path_sum(None, 0)
 0
 >>> BinaryTreePathSum().path_sum(tree, 0)
 0

 The second tree looks like this
 0
 / \
 5 5

 >>> tree2 = Node(0)
 >>> tree2.left = Node(5)
 >>> tree2.right = Node(15)

 >>> BinaryTreePathSum().path_sum(tree2, 5)
 2
 >>> BinaryTreePathSum().path_sum(tree2, -1)
 0
 >>> BinaryTreePathSum().path_sum(tree2, 0)
 1

 """

 target: int
Expand Down

88 changes: 88 additions & 0 deletions
data_structures/binary_tree/invert_binary_tree.py

View file


Open in desktop

Original file line numberDiff line numberDiff line change
@@ -0,0 +1,88 @@
"""
Given the root of a binary tree, invert the tree and return its root.

Leetcode: https://leetcode.com/problems/invert-binary-tree/description/

If n is the number of nodes in the tree, then:
Time complexity: O(n) as every subtree needs to be mirrored, we visit each node once.

Space complexity: O(h) where h is the height of the tree. This recursive algorithm 
uses the space of the stack which can grow to the height of the binary tree.
The space complexity will be O(n log n) for a binary tree and O(n) for a skewed tree.
"""

from __future__ import annotations 
from dataclasses import dataclass 

@dataclass
class TreeNode:
 """
 A TreeNode has a data variable and pointers to TreeNode objects for its left and right children.
 """

 def __init__(self, data: int) -> None:
 self.data = data 
 self.left: TreeNode | None = None 
 self.right: TreeNode | None = None 


class MirrorBinaryTree:
 def invert_binary_tree(self, root : TreeNode):
 """
 Invert a binary tree and return the new root.

 Returns the root of the mirrored binary tree.

 >>> tree = TreeNode(0)
 >>> tree.left = TreeNode(10)
 >>> tree.right = TreeNode(20)
 >>> result_tree = MirrorBinaryTree().invert_binary_tree(tree)
 >>> print_preorder(result_tree)
 0
 20
 10
 >>> tree2 = TreeNode(9)
 >>> result_tree2 = MirrorBinaryTree().invert_binary_tree(tree2)
 >>> print_preorder(result_tree2)
 9
 """ 

 if not root:
 return None 

 if root.left:
 self.invert_binary_tree(root.left)

 if root.right:
 self.invert_binary_tree(root.right)

 root.left, root.right = root.right, root.left 

 return root

def print_preorder(root: TreeNode | None) -> None:
 """
 Print pre-order traversal of the tree .

 >>> root = TreeNode(1)
 >>> root.left = TreeNode(2)
 >>> root.right = TreeNode(3)
 >>> print_preorder(root)
 1
 2
 3
 >>> print_preorder(root.right)
 3
 """
 if not root:
 return None 
 if root:
 print(root.data)
 print_preorder(root.left)
 print_preorder(root.right)


if __name__ == "__main__":
 import doctest 
 doctest.testmod()