LeetCode || Validate Binary Search Tree

最新推荐文章于 2021-01-26 17:17:22 发布

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#LeetCode #Depth-First Search #InOrder #Tree

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本文介绍了一种通过中序遍历来验证给定二叉树是否为有效的二叉搜索树的方法。有效二叉搜索树需满足左子树的所有节点小于根节点，右子树的所有节点大于根节点，且左右子树也必须是有效的二叉搜索树。

Validate Binary Search Tree

题目链接耗时:16ms

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * 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:

    //中序遍历的实现
    void GetInOrder(TreeNode* root, vector<int>& v)
    {
        if(root == NULL)
            return;
        
        if(root->left!=NULL)
        {
            GetInOrder(root->left, v);
        }
        v.push_back(root->val);
        if(root->right != NULL)
            GetInOrder(root->right, v);
    }
    
    bool isValidBST(TreeNode* root) {
        //错误的做法：运用递归来实现遍历，避免出现【10， 5， 15， null, null, 6， 20】此类树的结构从而出现错误的判断。
        //正确的做法: 利用中序遍历得到的是排好序的结构。
        
        vector<int> v;
        GetInOrder(root, v);
        
        for(int i = 1;i<v.size();i++)
        {
            if(v[i] <= v[i-1])
                return false;
        }
        
        return true;
    }
};