[LeetCode] 129: Validate Binary Search Tree

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本文介绍了一种检查给定二叉树是否为有效二叉搜索树的方法。通过递归算法确保每个节点的值大于其左子树的所有节点且小于右子树的所有节点，并保证左右子树也遵循同样的规则。

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[Problem]

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

[Solution]

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
	bool isValidBST(TreeNode *root, int leftVal, int rightVal){
		// null node
		if(NULL == root)return true;
		return root->val > leftVal && root->val < rightVal && isValidBST(root->left, leftVal, root->val) && isValidBST(root->right, root->val, rightVal);
	}
	// check if a tree is a BST
	bool isValidBST(TreeNode *root) {
		// Note: The Solution object is instantiated only once and is reused by each test case.
		return isValidBST(root, INT_MIN, INT_MAX);
	}
};