leetcode之validate-binary-search-tree(判断二叉树是否是BST搜索二叉树)

最新推荐文章于 2021-04-01 22:32:58 发布

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最新推荐文章于 2021-04-01 22:32:58 发布

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分类专栏： C++学习记录文章标签： leetcode BST

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博客围绕LeetCode的validate - binary - search - tree题目展开，要求判断给定二叉树是否为BST搜索二叉树。介绍了题意，解题思路是对二叉树进行中序遍历得到数组，判断数组是否递增，还提及了二叉树序列化方式。

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leetcode之validate-binary-search-tree(判断二叉树是否是BST搜索二叉树)

题目

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.

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:
在这里插入图片描述
The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

题意

给定一个二叉树，判断这个二叉树是否是BST搜索二叉树。

解题思路

BST 左子树比根节点小，右子树比根节点大。直接将该二叉树进行中序遍历，输出数组，判断是否递增。

C++实现代码

/**
 * 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) {
        vector<int > v;
        inorder(root,v);
        int num = v.size();
        /////判断v是不是递增//////
        for(int i=1;i<num;i++){
            if(v[i-1]>=v[i]){
                return false;
            }
        }
        return true;
    };
    void inorder(TreeNode* root,vector<int> &v){
        if(root == nullptr){
            return ;
        }
        inorder(root->left,v);
        v.push_back(root->val);
        inorder(root->right,v);
        
    }
};