Validate Binary Search Tree

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.

 判断是否是二叉搜索树有陷阱：http://blog.csdn.net/sgbfblog/article/details/7771096

C++实现代码如下，使用的是中序遍历是否为递增的序列来判断：

#include<iostream>

#include<new>

#include<vector>

#include<climits>

using namespace std;



//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==NULL)

            return true;

        vector<int> rvec;

        preorder(root,rvec);

        size_t i;

        int max;

        if(rvec.size())

            max=rvec[0];

        for(i=1;i<rvec.size();i++)

        {

            if(rvec[i]>max)

            {

                max=rvec[i];

            }

            else

                return false;

        }

        return true;

    }

    void preorder(TreeNode *root,vector<int> &rvec)

    {

        if(root)

        {

            preorder(root->left,rvec);

            rvec.push_back(root->val);

            preorder(root->right,rvec);

        }

    }

    void createTree(TreeNode *&root)

    {

        int i;

        cin>>i;

        if(i!=0)

        {

            root=new TreeNode(i);

            if(root==NULL)

                return;

            createTree(root->left);

            createTree(root->right);

        }

    }

};



int main()

{

    Solution s;

    TreeNode *root;

    s.createTree(root);

    cout<<s.isValidBST(root)<<endl;

}

错误的方法：

#include<iostream>

#include<new>

#include<vector>

using namespace std;



//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==NULL)

            return true;

        if(root->left&&root->right)

            return (root->left->val<root->val)&&(root->val<root->right->val)&&isValidBST(root->left)&&isValidBST(root->right);

        else if(root->left)

            return (root->left->val<root->val)&&isValidBST(root->left);

        else if(root->right)

            return (root->val<root->right->val)&&isValidBST(root->right);

    }

    void createTree(TreeNode *&root)

    {

        int i;

        cin>>i;

        if(i!=0)

        {

            root=new TreeNode(i);

            if(root==NULL)

                return;

            createTree(root->left);

            createTree(root->right);

        }

    }

};



int main()

{

    Solution s;

    TreeNode *root;

    s.createTree(root);

    cout<<s.isValidBST(root)<<endl;

}