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98. Validate Binary Search Tree

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Description

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



Input: [2,1,3]

Output: true

Example 2:

    5

   / \

  1   4

     / \

    3   6



Input: [5,1,4,null,null,3,6]

Output: false

Explanation: The root node's value is 5 but its right child's value is 4.

Solutions

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    pre = None
    def isValidBST(self, root: TreeNode) -> bool:
        if not root:
            return True
        if not self.isValidBST(root.left):
            return False
        if self.pre is not None and self.pre >= root.val:
            return False
        self.pre = root.val
        if not self.isValidBST(root.right):
            return False
        return True

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Integer pre = null;
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        if (!isValidBST(root.left)) return false;
        if (pre != null && pre >= root.val) return false;
        pre = root.val;
        if (!isValidBST(root.right)) return false;
        return true;
    }
}

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