[LeetCode]Validate Binary Search Tree-优快云博客

本文介绍了一种判断给定二叉树是否为有效二叉搜索树的方法。通过两种思路进行阐述：一是检查每个节点的左右子树的最大最小值；二是采用中序遍历，确保遍历结果为递增序列。

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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.

鉴定一颗树是否是二叉搜索树。思路：

1）左子树最大点小于根节点，根节点小于右子树最小节点。

2）中序遍历是有序的。

思路1：

class Solution {
public:
    bool isValidBST(TreeNode* root) {
        if(root==NULL)
            return true;
        if((root->left==NULL||Max(root->left)<root->val)&&(root->right==NULL||Min(root->right)>root->val))
            return isValidBST(root->left)&&isValidBST(root->right);
        else
            return false;
    }
    int Max(TreeNode *temp){
        if(temp->right==NULL)
            return temp->val;
        else return Max(temp->right);
    }
     int Min(TreeNode *temp){
        if(temp->left==NULL)
            return temp->val;
        else return Min(temp->left);
    }

思路2：

class Solution {
public:
    bool isValidBST(TreeNode* root) {
        if(root==NULL)
            return true;
       vector<int> temp;
       search(root,temp);
       if(temp.size()==1)
            return true;
       for(int i=0; i<temp.size()-1; ++i){
           if(temp[i]>=temp[i+1])
                return false;
       }
       return true;
    }
    void search(TreeNode* root,vector<int> &temp){
        if(root->left == NULL&&root->right==NULL){
            temp.push_back(root->val);
            return;
        }
        if(root->left)
            search(root->left,temp);
        temp.push_back(root->val);
        if(root->right)
            search(root->right,temp);
    }

};