LeetCode 0098 Validate Binary Search Tree【二叉搜索树】

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.

Example 1:

    2
   / \
  1   3

Input: [2,1,3]
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6

Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

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题意

判断一棵二叉树是不是二叉搜索树

思路1

• 中序遍历二叉树，记录遍历结果
• 看中序遍历序列是否是递增的
• 注意还需要去重 res.erase(unique(res.begin(), res.end()),res.end());

代码1

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isValidBST(TreeNode* root) {
        vector<int> ans;
        inorder(root, ans);
        vector<int> res(ans);
        
        sort(res.begin(), res.end());
        // 去重
        res.erase(unique(res.begin(), res.end()),res.end());
        return res == ans;
    }
    
    void inorder(TreeNode *node, vector<int> &ans){
        if(node == NULL) return;
        inorder(node->left, ans);
        ans.push_back(node->val);
        inorder(node->right, ans);
    }
};

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思路2

在上面做法的基础上不需要排序，只需要检查前一个元素是不是大于等于后面的元素即可，若满足，则不是一颗二叉搜索树，直接退出即可。

代码2

/**
 * Definition for a binary tree node.
 * 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> ans;
        inorder(root, ans);
        bool flag = true;
        for(int i = 0; i < ans.size() - 1; i++)
            if(ans[i] >= ans[i+1])
            {
                flag = false;
                break;
            }
        return flag;
    }
    
    void inorder(TreeNode *node, vector<int> &ans){
        if(node == NULL) return;
        inorder(node->left, ans);
        ans.push_back(node->val);
        inorder(node->right, ans);
    }
};