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问题描述:
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.
示例:
2/ \1 3Binary tree
[2,1,3]
, return true.
1/ \2 3Binary tree
[1,2,3]
, return false.
问题分析:
根据二叉搜索树而言,深度优先遍历后得到一个递增的数组。到时验证这个数组的合理性就能判断树是否为BST了。
过程详见代码:
/*** 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:bool isValidBST(TreeNode* root) {vector<int> res;bl(root, res);for (int i = 1; i < res.size(); i++){if (res[i - 1] >= res[i]) return false;}return true;}void bl(TreeNode* root,vector<int> &res){if (root == NULL) return;bl(root->left, res);res.push_back(root->val);bl(root->right, res);}
};
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