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Merge Two Binary Trees

LeetCode 617 | Difficulty: Easy​

Easy

Problem Description​

You are given two binary trees root1 and root2.

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 the two trees 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 the new tree.

Return the merged tree.

Note: The merging process must start from the root nodes of both trees.

Example 1:

Input: root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]
Output: [3,4,5,5,4,null,7]

Example 2:

Input: root1 = [1], root2 = [1,2]
Output: [2,2]

Constraints:

- The number of nodes in both trees is in the range `[0, 2000]`.

- `-10^4 <= Node.val <= 10^4`

Topics: Tree, Depth-First Search, Breadth-First Search, Binary Tree


Approach​

Tree DFS​

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left β†’ go right β†’ combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

Tree BFS (Level-Order)​

Use a queue to process the tree level by level. At each level, process all nodes in the queue, then add their children. Track the level size to know when one level ends and the next begins.

When to use

Level-order traversal, level-based aggregation, right/left side view.


Solutions​

Solution 1: C# (Best: 148 ms)​

MetricValue
Runtime148 ms
Memory25.9 MB
Date2019-02-11
Solution
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode MergeTrees(TreeNode t1, TreeNode t2) {
if(t1==null && t2==null) return null;
if(t1==null) return t2;
if(t2==null) return t1;
t1.val = t2.val+t1.val;
t1.left = MergeTrees(t1.left, t2.left);
t1.right = MergeTrees(t1.right, t2.right);
return t1;
}
}

Complexity Analysis​

ApproachTimeSpace
Tree Traversal$O(n)$$O(h)$

Interview Tips​

Key Points
  • Start by clarifying edge cases: empty input, single element, all duplicates.
  • Consider: "What information do I need from each subtree?" β€” this defines your recursive return value.