leetcode 653. Two Sum IV - Input is a BST 中序遍历 + 深度优先遍历DFS

Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.

Example 1:

Input:

5

/ \

3 6

/ \ \

2 4 7

Target = 9

Output: True

Example 2:

Input:

5

/ \

3 6

/ \ \

2 4 7

Target = 28

Output: False

本题题意很简单，直接做一个中序遍历，存储到set中，然后做一次查询即可

这个是必须要注意的：注意set做查询的时候要把当前元素删除，然后查询之后再插入

代码如下：

#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <string>
#include <climits>
#include <algorithm>
#include <sstream>
#include <functional>
#include <bitset>
#include <numeric>
#include <cmath>
#include <regex>

using namespace std;

/*
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
*/

class Solution
{
public:
set<int> all;
bool findTarget(TreeNode* root, int k)
{
dfs(root);
for (auto i : all)
{
all.erase(i);
if (all.find(k - i) != all.end())
return true;
all.insert(i);
}
return false;
}

void dfs(TreeNode* root)
{
if (root == NULL)
return;
else
{
dfs(root->left);
all.insert(root->val);
dfs(root->right);
}
}
};