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.

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

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}".

思路：判断二叉树是否满足二叉搜索树，可以使用中序遍历的方法，将二叉树遍历一遍，然后判断这个中序遍历是否是递增的，如果不是则返回false，如果不是则返回true；

/**

 * 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);

    }

};