# LeetCode 98 Validate Binary Search Tree DFS

2022-05-16 06:31:25  阅读：37  来源： 互联网

Given the `root` of a binary tree, determine if it is a valid binary search tree (BST).

A valid 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.

### Solution

\(BST\) 满足对于中序遍历，得到的节点值为递增序列。所以先中序遍历以后，检查序列是否严格递增即可：

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
private:
vector<int> node;

void dfs(TreeNode* root){
if(!root) return;
dfs(root->left); node.push_back(root->val); dfs(root->right);
}
public:
bool isValidBST(TreeNode* root) {
if(root==NULL) return true;
dfs(root);
int n = node.size();
for(int i=1;i<n;i++){
if(node[i-1]>=node[i])return false;
}
return true;
}

};

``````