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);
    }
};