leetcode98---Validate Binary Search Tree

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.

confused what “{1,#,2,3}” means?

OJ’s Binary Tree Serialization:

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}”.

问题求解：

方法一：

/**
* 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) {
return JudgeBST(root,LONG_LONG_MIN,LONG_LONG_MAX);
}
bool JudgeBST(TreeNode *root,long long min,long long max)
{
if(root == NULL) return true;
if(root->val<=min || root->val>=max) return false;
return JudgeBST(root->left, min, root->val)&&JudgeBST(root->right, root->val, max);
}
};

方法二：

最坏情况复杂度为O（n^2)

class Solution {
public:
bool isValidBST(TreeNode* root) {
if(!root) return true;
if(root->left)
{//左子树所有结点的值均小于它的根结点的值
//因此左子树最大节点的也应小于根结点的值
TreeNode *p=root->left;
while(p->right)
{//一直向右子树遍历
p=p->right;
}//找到最大值
if(p->val >= root->val)
{
return false;
}
}
if(root->right)
{//右子树所有结点的值均大于它的根结点的值
//因此右子树最小节点的也应大于根结点的值
TreeNode *p=root->right;
while(p->left)
{//一直向左子树遍历
p=p->left;
}//找到最小值
if(p->val <= root->val)
{
return false;
}
}
return isValidBST(root->left)&&isValidBST(root->right);
}

};