标签：des style blog class code java
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
confused
what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1
/ 2 3
/
4
5
The above
binary tree is serialized
as "{1,2,3,#,#,4,#,#,5}".
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void InOrder(TreeNode *root,vector<int> &result) { if(root==NULL) return; InOrder(root->left,result); result.push_back(root->val); InOrder(root->right,result); } bool isValidBST(TreeNode *root) { vector<int> result; result.clear(); InOrder(root,result); int n=result.size(); for(int i=1;i<n;i++) { if(result[i]<=result[i-1]) return false; } return true; } };
第二种思路：递归判断这个二叉树。首先定义左右边界，判断该节点是否处于这个范围之内，然后递归调用左子树，左右边界为min和root->val，递归调用右子树，左右边界为root->val和max，如果这两者都为真，则返回true，否则返回false。
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool check(TreeNode *root,int min,int max) { if(root==NULL) return true; if(root->val>min && root->val<max) { return check(root->left,min,root->val) && check(root->right,root->val,max); } else return false; } bool isValidBST(TreeNode *root) { int min=INT_MIN; int max=INT_MAX; return check(root,min,max); } };
Validate Binary Search Tree,布布扣,bubuko.com
标签：des style blog class code java
原文地址：http://www.cnblogs.com/awy-blog/p/3703750.html
