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.
Solution:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isValidBST(TreeNode root) {
if(root==null || (root.left==null && root.right==null)) return true;
if(root.left!=null && root.right==null){
TreeNode largest = root.left;
while(largest.right!=null) largest = largest.right;
return largest.val<root.val && isValidBST(root.left);
} else if(root.left==null && root.right!=null){
TreeNode smallest = root.right;
while(smallest.left!=null) smallest = smallest.left;
return smallest.val>root.val && isValidBST(root.right);
}
TreeNode largest = root.left;
while(largest.right!=null) largest = largest.right;
TreeNode smallest = root.right;
while(smallest.left!=null) smallest = smallest.left;
return largest.val<root.val && smallest.val>root.val && isValidBST(root.left) && isValidBST(root.right);
}
}