leetcode102：二叉树的层序遍历

[sisterAn]

给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。 

示例：
二叉树：[3,9,20,null,null,15,7] ,

 3
 / \
 9 20
 / \
 15 7

返回其层次遍历结果：

[
 [3],
 [9,20],
 [15,7]
]

[plane-hjh]

这道题和二叉树的层次遍历相似，只需要把 reverse() 去除就可以了

广度优先遍历

var levelOrder = function(root) {
 if(!root) return []
 let res = [], 
 queue = [root]
 while(queue.length) {
 let curr = [],
 temp = []
 while(queue.length) {
 let node = queue.shift()
 curr.push(node.val)
 if(node.left) temp.push(node.left)
 if(node.right) temp.push(node.right)
 }
 res.push(curr)
 queue = temp
 }
 return res
};

深度优先遍历

var levelOrder = function(root) {
 const res = []
 var dep = function (node, depth){
 if(!node) return
 res[depth] = res[depth]||[]
 res[depth].push(node.val)
 dep(node.left, depth + 1)
 dep(node.right, depth + 1)
 }
 dep(root, 0)
 return res
};

[AnranS]

var levelOrder = function(root) {
 let res = [], queue = [];
 if(!root) return res;
 queue.push(root);

 while(queue.length) {
 let len = queue.length;
 res.push([]);
 for(let i = 0;i<len;i++) {
 let node = queue.shift();
 res[res.length - 1].push(node.val);
 if(node.left) queue.push(node.left);
 if(node.right) queue.push(node.right);
 }
 }
 return res;
};

[lewisYe]

使用广度优先遍历 然后再对数据进行分层

var levelOrder = function(root) {
 if(!root || root.val == null) return []
 let queue = []
 let res = []
 queue.push(root)
 while(queue.length){
 let len = queue.length
 let curr = []
 for(var i =0;i<len;i++){
 let node = queue.shift()
 curr.push(node.val)
 if(node.left)queue.push(node.left)
 if(node.right)queue.push(node.right)
 }
 res.push(curr)
 }
 return res
};