Question

Formatted question description: https://leetcode.ca/all/1214.html

 Given two binary search trees,
 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:
   2     1
  / \   / \
 1   4 0   3

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

 @tag-array

Algorithm

Get two ordered list, and 2 pointers to move forward for each.

Code

import java.util.ArrayList;
import java.util.List;


public class Two_Sum_BSTs {

    // Time: O(M+N)
    // Space: O(logmax(M,N)) in the best case; O(max(M,N)) in the worst case.
    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {

        // Convert to list
        List<Integer> list1 = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();
        listByInorder(root1, list1);
        listByInorder(root2, list2);

        // Two Pointers
        int lo = 0, hi = list2.size() - 1;
        while (lo < list1.size() && hi >= 0) { // notice the condition here
            int sum = list1.get(lo) + list2.get(hi);
            if (sum == target) {
                return true;
            } else if (sum < target) {
                ++lo;
            } else {
                --hi;
            }
        }
        return false;
    }

    private void listByInorder(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        listByInorder(root.left, list);
        list.add(root.val);
        listByInorder(root.right, list);
    }
}

############

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean twoSumBSTs(TreeNode root1, TreeNode root2, int target) {
        List<Integer> vals1 = new ArrayList<>();
        List<Integer> vals2 = new ArrayList<>();
        inorder(root1, vals1);
        inorder(root2, vals2);
        int i = 0, j = vals2.size() - 1;
        while (i < vals1.size() && j >= 0) {
            int s = vals1.get(i) + vals2.get(j);
            if (s == target) {
                return true;
            }
            if (s < target) {
                ++i;
            } else {
                --j;
            }
        }
        return false;
    }

    private void inorder(TreeNode root, List<Integer> vals) {
        if (root != null) {
            inorder(root.left, vals);
            vals.add(root.val);
            inorder(root.right, vals);
        }
    }
}

// OJ: https://leetcode.com/problems/two-sum-bsts/
// 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;
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def twoSumBSTs(self, root1: TreeNode, root2: TreeNode, target: int) -> bool:
        vals1, vals2 = [], []

        def inorder(root, vals):
            if root:
                inorder(root.left, vals)
                vals.append(root.val)
                inorder(root.right, vals)

        inorder(root1, vals1)
        inorder(root2, vals2)

        i, j = 0, len(vals2) - 1
        while i < len(vals1) and j >= 0:
            if vals1[i] + vals2[j] == target:
                return True
            if vals1[i] + vals2[j] < target:
                i += 1
            else:
                j -= 1
        return False

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func twoSumBSTs(root1 *TreeNode, root2 *TreeNode, target int) bool {
	vals1 := inorder(root1)
	vals2 := inorder(root2)
	i, j := 0, len(vals2)-1
	for i < len(vals1) && j >= 0 {
		s := vals1[i] + vals2[j]
		if s == target {
			return true
		}
		if s < target {
			i++
		} else {
			j--
		}
	}
	return false
}

func inorder(root *TreeNode) []int {
	if root == nil {
		return nil
	}
	left := inorder(root.Left)
	right := inorder(root.Right)
	return append(append(left, root.Val), right...)
}

1214 - Two Sum BSTs

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