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/**
* Given a binary tree, determine if it is a valid binary search tree (BST).
*
* Assume a BST is defined as follows:
*
* The left subtree of a node contains only nodes with keys less than the node's key.
* The right subtree of a node contains only nodes with keys greater than the node's key.
* Both the left and right subtrees must also be binary search trees.
*
* Example 1:
* 2
* / \
* 1 3
* Binary tree [2,1,3], return true.
*
* Example 2:
* 1
* / \
* 2 3
* Binary tree [1,2,3], return false.
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
publicclassValidateBinarySearchTree98 {
publicbooleanisValidBST(TreeNoderoot) {
returnhelper(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
privatebooleanhelper(TreeNoderoot, longmin, longmax) {
if (root == null) returntrue;
if (root.val <= min || root.val >= max) returnfalse;
returnhelper(root.left, min, root.val) && helper(root.right, root.val, max);
}
publicbooleanisValidBST2(TreeNoderoot) {
if (root == null) returntrue;
returnisValidBST(root, newint[]{0, 0});
}
privatebooleanisValidBST(TreeNoderoot, int[] bounds) {
bounds[0] = root.val;
bounds[1] = root.val;
if (root.left == null && root.right == null) {
returntrue;
}
if (root.left != null) {
int[] leftBounds = newint[]{0, 0};
booleanleft = isValidBST(root.left, leftBounds);
if (!left || leftBounds[1] >= root.val) returnfalse;
bounds[0] = leftBounds[0];
}
if (root.right != null) {
int[] rightBounds = newint[]{0, 0};
booleanright = isValidBST(root.right, rightBounds);
if (!right || rightBounds[0] <= root.val) returnfalse;
bounds[1] = rightBounds[1];
}
returntrue;
}
/**
* https://discuss.leetcode.com/topic/46016/learn-one-iterative-inorder-traversal-apply-it-to-multiple-tree-questions-java-solution
*/
publicbooleanisValidBST3(TreeNoderoot) {
if (root == null) returntrue;
Stack<TreeNode> stack = newStack<>();
TreeNodepre = null;
while (root != null || !stack.isEmpty()) {
while (root != null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
if(pre != null && root.val <= pre.val) returnfalse;
pre = root;
root = root.right;
}
returntrue;
}
}