README.md

617. Merge Two Binary Trees

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.

Solutions (C)

1. DFS

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* mergeTrees(struct TreeNode* t1, struct TreeNode* t2){
        if(t1 == NULL || t2 == NULL)
            return t1 == NULL ? t2 : t1;
        t1 -> val += t2 -> val;
        t1 -> left = mergeTrees(t1 -> left, t2 -> left);
        t1 -> right = mergeTrees(t1 -> right, t2 -> right);

        return t1;
}

Solutions (C++)

1. DFS

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if(t1 == nullptr || t2 == nullptr)
            return t1 == nullptr ? t2 : t1;
        t1 -> val += t2 -> val;
        t1 -> left = mergeTrees(t1 -> left, t2 -> left);
        t1 -> right = mergeTrees(t1 -> right, t2 -> right);

        return t1;
    }
};

Solutions (Java)

1. DFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if(t1 == null || t2 == null)
            return t1 == null ? t2 : t1;
        t1.val += t2.val;
        t1.left = mergeTrees(t1.left, t2.left);
        t1.right = mergeTrees(t1.right, t2.right);
        return t1;
    }
}

Solutions (Kotlin)

1.DFS

/**
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    fun mergeTrees(t1: TreeNode?, t2: TreeNode?): TreeNode? {
        // return if(t1 == null) t2 else t1
        if (t1 == null) {
            return t2
        }
        if (t2 == null) {
            return t1
        }
        t1.`val` = t1.`val`+ t2.`val`
        t1.left = mergeTrees(t1.left, t2.left)
        t1.right = mergeTrees(t1.right, t2.right)

        return t1
    }
}

Solutions(Python)

1.DFS

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def mergeTrees(self, t1: TreeNode, t2: TreeNode) -> TreeNode:
        if not t1 or not t2: 
            return t1 or t2
        t1.val = t1.val + t2.val
        t1.left = self.mergeTrees(t1.left, t2.left)
        t1.right = self.mergeTrees(t1.right, t2.right)
        return t1