leetcode第六周解题总结（98）

98. Validate Binary Search Tree

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.

题意解析：

判断是否为合法的二叉搜索树，三个条件：

一个节点的左子树只含有比当前节点小的值

一个节点的右子树只含有比当前节点大的值

左子树和右子树也必须是二叉搜索树

解题思路：

中序遍历。

/**
* 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:
vector<int> a;
bool isValidBST(TreeNode* root) {
if(root == NULL) return true;
if(root -> left == NULL && root->right == NULL) return true;
inorder(root);
for(int i = 0; i < a.size() - 1; i ++) {
if (a[i] >= a[i + 1]) {
return false;
}
}
return true;
}

vector<int> inorder(TreeNode* root) {
if(root->left != NULL) {
inorder(root->left);
}
a.push_back(root->val);
if(root->right != NULL) {
inorder(root->right);
}
return a;
}
};