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

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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 thanthe 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;
    }
};