【leetcode】Validate Binary Search Tree

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.

 

思路一：利用递归的思想，任何一个节点必须要在min和max范围内

 

min和max根据是左子树还是右子树来确定

 

如果是左子树：

node->left<node

 

如果是右子树：

node->right>node

 

 

递归条件：node->val在min和max范围内，node->left要在min,node->val范围内，node->right要在node->val,max范围内

node->val<max&&node->val>min&&testValid(node->left,node->val,min)&&testValid(node->right,max,node->val); 

 

 

 

/**
 * 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==NULL&&root->right==NULL)
        {
            return true;
        }
       
        //注意如果测试用例含有INT_MAX则，必须采用long long int 才能避免出错
        return testValid(root,(long long int)INT_MAX+1,(long long int)INT_MIN-1);
    }
   
   
    bool testValid(TreeNode *node,long long int max, long long int min)
    {
        if(node==NULL)
        {
            return true;
        }
       
       
        return node->val<max&&node->val>min&&testValid(node->left,node->val,min)&&testValid(node->right,max,node->val);
       
    }
};

 

 

 

第二种方法，利用 二叉排序树 中序遍历 结果为递增序列的性质

 

 

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> ret_v;
    bool isValidBST(TreeNode *root) {
       
        if(root==NULL)
        {
            return true;
        }
       
        ret_v.clear();
        inOrderTraversal(root);
       
        //判断是否递增
        for(int i=0;i<ret_v.size()-1;i++)
        {
            if(ret_v[i]>=ret_v[i+1])
            {
                return false;
            }
        }
        return true;
    }
   
    //中序遍历，记录下数值
    void inOrderTraversal(TreeNode *root)
    {
        if(root==NULL)
        {
            return;
        }
       
        inOrderTraversal(root->left);
        ret_v.push_back(root->val);
        inOrderTraversal(root->right);
    }
};

 

 

 

第三种：直接在中序遍历中判断：

 

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
   
    int ret;
    bool flag;
    bool isFirstNode;
 
    bool isValidBST(TreeNode *root) {
       
        flag=true;
        //判断是不是第一个被访问的节点
        isFirstNode=true;
 
        inOrderTraversal(root);
        return flag;
    }
   
    void inOrderTraversal(TreeNode *root)
    {
        if(root==NULL)
        {
            return;
        }
       
        if(!flag)
        {
            return;
        }
       
        inOrderTraversal(root->left);
       
        //一旦发现不符合升序，则不是二叉排序树
        if(!isFirstNode&&ret>=root->val)
        {
            flag=false;
            return;
        }
 
        ret=root->val;
        isFirstNode=false;
       
        inOrderTraversal(root->right);
    }
   
 
};