653. Two Sum IV - Input is a BST

 

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

 

这道题又是一道2sum的变种题,博主一直强调,平生不识TwoSum,刷尽LeetCode也枉然!只要是两数之和的题,一定要记得先尝试用HashSet来做,这道题只不过是把数组变成了一棵二叉树而已,换汤不换药,我们遍历二叉树就行,然后用一个HashSet,在递归函数函数中,如果node为空,返回false。如果k减去当前结点值在HashSet中存在,直接返回true;否则就将当前结点值加入HashSet,然后对左右子结点分别调用递归函数并且或起来返回即可,参见代码如下:

 

解法一:

class Solution {
public:
    bool findTarget(TreeNode* root, int k) {
        unordered_set<int> st;
        return helper(root, k, st);
    }
    bool helper(TreeNode* node, int k, unordered_set<int>& st) {
        if (!node) return false;
        if (st.count(k - node->val)) return true;
        st.insert(node->val);
        return helper(node->left, k, st) || helper(node->right, k, st);
    }
};

 

我们也可以用层序遍历来做,这样就是迭代的写法了,但是利用HashSet的精髓还是没变的,参见代码如下:

 

解法二:

class Solution {
public:
    bool findTarget(TreeNode* root, int k) {
        if (!root) return false;
        unordered_set<int> st;
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            auto t = q.front(); q.pop();
            if (st.count(k - t->val)) return true;
            st.insert(t->val);
            if (t->left) q.push(t->left);
            if (t->right) q.push(t->right);
        }
        return false;
    }
};

 

由于输入是一棵二叉搜索树,那么我们可以先用中序遍历得到一个有序数组,然后在有序数组中找两数之和就很简单了,直接用双指针进行遍历即可,参见代码如下:

 

解法三:

class Solution {
public:
    bool findTarget(TreeNode* root, int k) {
        vector<int> nums;
        inorder(root, nums);
        for (int i = 0, j = (int)nums.size() - 1; i < j;) {
            if (nums[i] + nums[j] == k) return true;
            (nums[i] + nums[j] < k) ? ++i : --j;
        }
        return false;
    }
    void inorder(TreeNode* node, vector<int>& nums) {
        if (!node) return;
        inorder(node->left, nums);
        nums.push_back(node->val);
        inorder(node->right, nums);
    }
};

 

类似题目:

Two Sum III - Data structure design

Two Sum II - Input array is sorted

Two Sum

 

参考资料:

https://leetcode.com/problems/two-sum-iv-input-is-a-bst/

https://leetcode.com/problems/two-sum-iv-input-is-a-bst/discuss/106090/my-c-python-solution

https://leetcode.com/problems/two-sum-iv-input-is-a-bst/discuss/106059/JavaC%2B%2B-Three-simple-methods-choose-one-you-like

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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