Leetcode 98. Validate Binary Search Tree-优快云博客

本文介绍了一种判断二叉树是否为有效的二叉搜索树的方法。通过对二叉搜索树进行中序遍历，并检查遍历过程中节点值是否严格递增，从而确定其有效性。文章提供了一个基于C++实现的示例代码。

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Leetcode 98. Validate Binary Search Tree

题目描述

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

Binary tree [2,1,3], return true.

Example 2:
    1
   / \
  2   3

Binary tree [1,2,3], return false.

输入：树
输出：该树是否为二叉搜索树

思路

利用二叉搜索树的性质，其中序遍历必为顺序增加。通过DFS对其进行中序遍历，若存在一个节点不符合严格递增，则提早结束DFS，并返回false。时间复杂度 $O(n)$

代码

/**
 * 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:
    vector<int> v;
    bool flag;
    void dfs(TreeNode* root){
        if(root==NULL) return ;
        if(root->left!=NULL)
            dfs(root->left);
        if(!v.empty()&&root->val <= v[v.size()-1]){
            flag = false;
            return ;
        }else{
            v.push_back(root->val);
        }
        if(root->right!=NULL)
            dfs(root->right);

        return ;
    }
    bool isValidBST(TreeNode* root) {
        if(root==NULL) return true;
        flag = true;
        dfs(root);
        return flag;
    }
};