标签:des style blog class code c
Problem Description:
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 5The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}"
.
解法一:
二叉搜索树定义为或者是一棵空树,或者是具有下列性质的二叉树: 若它的左子树不空,则左子树上所有结点的值均小于它的根结点的值; 若它的右子树不空,则右子树上所有结点的值均大于它的根结点的值; 它的左、右子树也分别为二叉排序树。
因此直接采用遍历判断的方法,每次找到该节点左子树的最大值和右子树的最小值与当前节点比较,不满足条件则不是,满足再判断左右子树是否都满足条件。该解法最坏情况复杂度为O(n^2)。(最坏情况为所有结点都在一边的时候)
代码如下:
/** * 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 isValidBST(TreeNode *root) { if(!root) return true; if(root->left) { TreeNode *p=root->left; while(p->right) p=p->right; if(p->val>=root->val) return false; } if(root->right) { TreeNode *p=root->right; while(p->left) p=p->left; if(p->val<=root->val) return false; } return isValidBST(root->left)&&isValidBST(root->right); } };
利用二叉搜索树中序遍历有序递增的性质,在中序遍历的过程中判断左子树、当前节点和右子树是否满足条件,时间复杂度为O(n)。
代码如下:
/** * 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 IsBST(TreeNode* root, int &pre) { if (root == NULL) return true; if (IsBST(root->left,pre)) { if(root->val>pre)//判断当前结点值是否大于prev,此时prev为中序遍历时当前结点的前一个值。 { pre=root->val; return IsBST(root->right,pre); } else return false; } else return false; } bool isValidBST(TreeNode* root) { int pre=INT_MIN; return IsBST(root, pre); } };
Leetcode--Validate Binary Search Tree,布布扣,bubuko.com
Leetcode--Validate Binary Search Tree
标签:des style blog class code c
原文地址:http://blog.csdn.net/longhopefor/article/details/25687119