标签：查找树 strong 等价 turn 判断 代码 中序遍历 help root
Given a binary tree, determine if it is a valid binary search tree (BST). （Medium）
Assume a BST is defined as follows:
Example 1:
2 / 1 3
Binary tree [2,1,3], return true.
Example 2:
1 / 2 3
Binary tree [1,2,3], return false.
分析：
是否是二叉查找树和中序遍历是否以此递增是等价的，所以可以先做中序遍历，然后将结果数组做判断，注意判断空树。
代码：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 private: 12 vector<int> result; 13 void helper(TreeNode* root) { 14 if (root == nullptr) { 15 return; 16 } 17 helper(root -> left); 18 result.push_back(root -> val); 19 helper(root -> right); 20 } 21 public: 22 bool isValidBST(TreeNode* root) { 23 helper(root); 24 if (result.size() == 0) { 25 return true;; 26 } 27 for (int i = 0; i < result.size() - 1; ++i) { 28 if (result[i] >= result[i + 1]) { 29 return false; 30 } 31 } 32 return true; 33 } 34 };
LeetCode98 Validate Binary Search Tree
标签：查找树 strong 等价 turn 判断 代码 中序遍历 help root
原文地址：http://www.cnblogs.com/wangxiaobao/p/6002240.html
