Algorithm
Problem Name: 617. Merge Two Binary Trees
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]. -104 <= Node.val <= 104
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
if (root1 == null && root2 == null) {
return null;
}
int value = (root1 == null ? 0 : root1.val) + (root2 == null ? 0 : root2.val);
TreeNode root = new TreeNode(value);
root.left = mergeTrees((root1 == null ? null : root1.left), (root2 == null ? null : root2.left));
root.right = mergeTrees((root1 == null ? null : root1.right), (root2 == null ? null : root2.right));
return root;
}
}
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def mergeTrees(self, t1, t2):
"""
:type t1: TreeNode
:type t2: TreeNode
:rtype: TreeNode
"""
if t1 and t2:
root, root.left, root.right = TreeNode(t1.val + t2.val), self.mergeTrees(t1.left, t2.left), self.mergeTrees(t1.right, t2.right)
return root
else: return t1 or t2
Input
Output
#3 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0617_MergeTwoBinaryTrees
{
public TreeNode MergeTrees(TreeNode t1, TreeNode t2)
{
if (t1 == null) return t2;
if (t2 == null) return t1;
var newNode = new TreeNode(t1.val + t2.val);
newNode.left = MergeTrees(t1.left, t2.left);
newNode.right = MergeTrees(t1.right, t2.right);
return newNode;
}
}
}
Input
Output