【Leetcode】98. Validate Binary Search Tree（二叉树中序遍历）

r囧r小猫 2022-02-20 19:18 202阅读 0赞

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:**

Input:
        2
       / \
      1   3
    Output: true

**Example 2:**

5
       / \
      1   4
         / \
        3   6
    Output: false
    Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
                 is 5 but its right child's value is 4.

题目大意：

问二叉树是否可以满足左边都小于当前节点，右边都大于当前节点。注意存在重复数据。

解题思路：

二叉树中序遍历，判断数组是否为升序序列。

/**
     * 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 {
    private:
        vector<int> ans;
        void search(TreeNode* tmp){
            if(tmp->left == NULL && tmp->right == NULL){
                ans.push_back(tmp->val);
                return;
            }
            if(tmp->left != NULL){
                search(tmp->left);
            }
            ans.push_back(tmp->val);
            if(tmp ->right != NULL){
                search(tmp->right);
            }
            
        }
    public:
        bool isValidBST(TreeNode* root) {
            if(root==NULL){
                return true;
            }
            search(root);
            for(int i =1;i<ans.size();i++){
                if(ans[i-1] >= ans[i]){
                    return false;
                }
            }
            return true;
        }
    };