[leetcode] Validate Binary Search Tree-优快云博客

本文介绍如何通过递归和迭代方法验证给定的二叉树是否为有效的二叉搜索树，包括使用中序遍历来检查顺序和利用辅助数组记录节点值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

From : https://leetcode.com/problems/validate-binary-search-tree/

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.

特别例子不能过，但很有意思：

class Solution {
public:
    bool check(TreeNode *node, long long int minV, long long int maxV) {
        if(node == NULL || !node->left && !node->right) return true;
		return (node->val>minV) && (node->val<maxV) && check(node->left, minV, node->val) && check(node->right, node->val, maxV);
    }
    
    bool isValidBST(TreeNode *root) {
		return check(root, -2147483649, 2147483648);
    }
};

Solution 1:

/** Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    
    bool isValidBST(TreeNode *root) {
        if(!root) return true;
        int val = root->val;
        if(!checkLeft(root->left, val) || !checkRight(root->right, val)) return false;
        return isValidBST(root->left) && isValidBST(root->right);        
    }
    
    bool checkLeft(TreeNode *node, int &val) {
        if(!node) return true;
        if(node->val < val) {
            return checkLeft(node->left, val) && checkLeft(node->right, val);
        }
        return false;
    }
    bool checkRight(TreeNode *node, int &val) {
        if(!node) return true;
        if(node->val > val) {
            return checkRight(node->left, val) && checkRight(node->right, val);
        }
        return false;
    }
};

solution 2:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode *root) {
        if(!root) return true;
        vector<int> nums;
        insert(root, nums);
        for(int i=nums.size()-1; i>0; i--) {
            if(nums[i] <= nums[i-1]) return false;
        }
        return true;        
    }
    void insert(TreeNode *node, vector<int> &nums) {
        if(node) {
            insert(node->left, nums);
            nums.push_back(node->val);
            insert(node->right, nums);
        }
    }
};

Solution 3：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
	bool isValidBST(TreeNode *root) {
		if(!root) return true;
		int val;
		bool isInited = false;
		return check(root, val, isInited);
	}
	bool check(TreeNode *root, int &pre, bool &isInited) {
		if (root != NULL) {
            if (!check(root->left, pre, isInited)) return false;
            if (isInited && root->val <= pre) return false;
            pre = root->val;
			isInited = true;
            return check(root->right, pre, isInited);
        }
        return true;
	}
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private boolean inited = false;
    private int p = 0;
    public boolean isValidBST(TreeNode root) {
        if(root == null) {
            return true;
        }
        return inTravelValid(root);
    }
    private boolean inTravelValid(TreeNode r) {
        if(r == null) {
            return true;
        }
        if(!inTravelValid(r.left)) {
            return false;
        }
        if(inited && r.val <= p) {
            return false;
        }
        p = r.val;
        inited = true;
        return inTravelValid(r.right);
        
    }
}

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