Leetcode Study Day 42

Validate Binary Search Tree

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid 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:


Input: root = [2,1,3]
Output: true
Example 2:


Input: root = [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.
 

Constraints:

The number of nodes in the tree is in the range [1, 104].
-231 <= Node.val <= 231 - 1

My solution

My solution is initialise a NodeTree variable called prev. Then we use in-order traverse to traverse the tree. If the previous node value is larger or equal to the current node value, then we return false. Otherwise, we return true.

class Solution {
public:
    TreeNode* prev = nullptr;
    bool isValidBST(TreeNode* root) {
        if(!root) return true;
        if (!isValidBST(root -> left)) return false;
        if (prev != nullptr && prev -> val >= root -> val)
            return false;
        prev = root;
        return (isValidBST(root -> right));
    }
};