98. Validate Binary Search Tree

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98. Validate Binary Search Tree

Description

Difficulty: Medium

Related Topics: Tree, Depth-First Search, Binary Search Tree, Binary 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:

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

Example 2:

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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

Hints/Notes

  • 2023/08/26
  • We need max/min node to decide if one node is valid
  • Because the val is between MIN_INT and MAX_INT, so use a nullptr node is better
  • 0x3F’s solution(checked)

Solution

Language: C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        return isValidBST(root, nullptr, nullptr);
    }

    bool isValidBST(TreeNode* root, TreeNode* min, TreeNode* max) {
        if (root == nullptr) return true;
        if (min && root->val <= min->val) return false;
        if (max && root->val >= max->val) return false;
        return isValidBST(root->left, min, root)
            && isValidBST(root->right, root, max);
    }
};