1214

1214. Two Sum BSTs (Medium)

Given the roots of two binary search trees, root1 and root2, return true if and only if there is a node in the first tree and a node in the second tree whose values sum up to a given integer target.

 

Example 1:

Input: root1 = [2,1,4], root2 = [1,0,3], target = 5
Output: true
Explanation: 2 and 3 sum up to 5.

Example 2:

Input: root1 = [0,-10,10], root2 = [5,1,7,0,2], target = 18
Output: false

 

Constraints:

• The number of nodes in each tree is in the range [1, 5000].
• -109 <= Node.val, target <= 109

Companies:
Amazon

Related Topics:
Two Pointers, Binary Search, Stack, Tree, Depth-First Search, Binary Search Tree, Binary Tree

Similar Questions:

• Two Sum IV - Input is a BST (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/two-sum-bsts/
// Author: github.com/lzl124631x
// Time: O(A + B)
// Space: O(HA + HB)
class Solution {
    void pushNodes(stack<TreeNode*> &s, TreeNode *n, bool right = false) {
        for (; n; n = right ? n->right : n->left) s.push(n);
    }
public:
    bool twoSumBSTs(TreeNode* a, TreeNode* b, int target) {
        stack<TreeNode*> sa, sb;
        pushNodes(sa, a);
        pushNodes(sb, b, true);
        while (sa.size() && sb.size()) {
            int sum = sa.top()->val + sb.top()->val;
            if (sum == target) return true;
            if (sum < target) {
                auto n = sa.top();
                sa.pop();
                pushNodes(sa, n->right);
            } else {
                auto n = sb.top();
                sb.pop();
                pushNodes(sb, n->left, true);
            }
        }
        return false;
    }
};