Implement a function to check if a binary tree is a binary search tree.
Example 1:
Input: 2 / \ 1 3 Output: true
Example 2:
Input: 5 / \ 1 4 / \ 3 6 Output: false Explanation: Input: [5,1,4,null,null,3,6]. the value of root node is 5, but its right child has value 4.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
res, t = True, None
def isValidBST(self, root: TreeNode) -> bool:
self.isValid(root)
return self.res
def isValid(self, root):
if not root:
return
self.isValid(root.left)
if self.t is None or self.t < root.val:
self.t = root.val
else:
self.res = False
return
self.isValid(root.right)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private boolean res = true;
private Integer t = null;
public boolean isValidBST(TreeNode root) {
isValid(root);
return res;
}
private void isValid(TreeNode root) {
if (root == null) {
return;
}
isValid(root.left);
if (t == null || t < root.val) {
t = root.val;
} else {
res = false;
return;
}
isValid(root.right);
}
}