Merge Two Binary Trees, by LeetCode
Late night problem: https://leetcode.com/problems/merge-two-binary-trees/description/
The solution is to recursively do exactly what the description says: if both exist, add them up. Otherwise, return the one that does not exist. Code is below, cheers, Marcelo.
Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.
You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.
Example 1:
Input: Tree 1 Tree 2 1 2 / \ / \ 3 2 1 3 / \ \ 5 4 7 Output: Merged tree: 3 / \ 4 5 / \ \ 5 4 7
Note: The merging process must start from the root nodes of both trees.
The solution is to recursively do exactly what the description says: if both exist, add them up. Otherwise, return the one that does not exist. Code is below, cheers, Marcelo.
public class Solution
{
public TreeNode MergeTrees(TreeNode t1, TreeNode t2)
{
if (t1 == null && t2 == null) return null;
if (t1 == null)
{
TreeNode retVal = new TreeNode(t2.val);
retVal.left = MergeTrees(t1, t2.left);
retVal.right = MergeTrees(t1, t2.right);
return retVal;
}
else if (t2 == null)
{
TreeNode retVal = new TreeNode(t1.val);
retVal.left = MergeTrees(t1.left, t2);
retVal.right = MergeTrees(t1.right, t2);
return retVal;
}
else
{
TreeNode retVal = new TreeNode(t1.val + t2.val);
retVal.left = MergeTrees(t1.left, t2.left);
retVal.right = MergeTrees(t1.right, t2.right);
return retVal;
}
}
}
I absolutely love these recursive problems! I've decided to reuse as much of existing trees as possible, but asymptotically both time and space complexity is the same:
ReplyDeleteclass Solution {
public:
TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
if (t1 == nullptr && t2 == nullptr) return nullptr;
if (t1 == nullptr) return t2;
if (t2 == nullptr) return t1;
auto newTree = new TreeNode(t1->val + t2->val);
newTree->left = mergeTrees(t1->left, t2->left);
newTree->right = mergeTrees(t1->right, t2->right);
return newTree;
}
};