标签：
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
Example 1:
2 / 1 3Binary tree
[2,1,3],
return true.
Example 2:
1 / 2 3Binary tree
[1,2,3],
return false.
判断是否为有效的二叉搜索树
/**
* 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 { //易错点，不是左结点的值小于根节点的值，是左子树所有的结点小于根节点
public: //解法1
//即左<根<右，初始化时带入系统最大值和最小值，在递归过程中换成它们自己的节点值，用long代替int就是为了包括int的边界条件。
bool isValidBST(TreeNode* root) {
return helper(root,LONG_MIN,LONG_MAX);
}
bool helper(TreeNode* root,long left,long right)
{
if(root==NULL)
return true;
if(root->val<=left||root->val>=right)
return false;
return helper(root->left,left,root->val)&&helper(root->right,root->val,right);
}
};/**
* 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 { //解法2，中序遍历,左根右
public:
vector<int> res;
bool isValidBST(TreeNode* root) {
if(root==NULL)return true;
inorder(root);
for(int i=0;i<res.size()-1;i++)
if(res[i]>=res[i+1])
return false;
return true;
}
void inorder(TreeNode* root)
{
stack<TreeNode*> s;
while(root!=NULL||!s.empty())
{
if(root!=NULL)
{
s.push(root);
root=root->left;
continue;
}
else
{
root=s.top();
s.pop();
res.push_back(root->val);
root=root->right;
}
}
}
};leetcode No98. Validate Binary Search Tree
标签：
原文地址：http://blog.csdn.net/u011391629/article/details/52227641
