題目鏈接
tag:
- Medium；
question：
??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.
思路：
??驗證二叉搜索樹有很多種解法团秽，可以利用它本身的性質(zhì)來做走触，即左<根<右晦譬，可以通過利用中序遍歷結(jié)果為有序數(shù)列來做，代碼如下：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
// Recursion
class Solution {
public:
bool isValidBST(TreeNode* root) {
if (!root) return true;
vector<int> vals;
inorder(root, vals);
for (int i = 0; i < vals.size() - 1; ++i) {
if (vals[i] >= vals[i + 1]) return false;
}
return true;
}
void inorder(TreeNode* root, vector<int>& vals) {
if (!root) return;
inorder(root->left, vals);
vals.push_back(root->val);
inorder(root->right, vals);
}
};
