Merge Two Binary Trees

• LeetCode: Merge Two Binary Tress
• Problem:

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.

• Note:

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

• Example:

Input: 
    Tree 1                     Tree 2                  
          1                         2                             
         / \                       / \                            
        3   2                     1   3                        
       /                           \   \                      
      5                             4   7                  
Output: 
Merged tree:
         3
        / \
       4   5
      / \   \ 
     5   4   7

• Analysis:

According to problem description, both trees are merged from both of root nodes.

We can use recursive solution to visit two trees at the same corresponding position nodes.

Recursion:

 mergedTree.val = tree1.val + tree2.val;

 mergedTree.leftTree = merge(tree1.left, tree2.left)

 mergedTree.rightTree = merge(tree2.right, tree2.right)

 return mergedTree

• Code - Java

/**
 * 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 null;
        }

        if (t1 == null) {
            return t2;
        }

        if (t2 == null) {
            return t1;
        }

        int val = t1.val + t2.val;
        TreeNode node = new TreeNode(val);
        node.left = mergeTrees(t1.left, t2.left);
        node.right = mergeTrees(t1.right, t2.right);
        return node;
    }
}

• Code - C++

/**
 * 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 == NULL && t2 == NULL) {
            return NULL;
        }

        if (t1 == NULL) {
            return t2;
        }

        if (t2 == NULL) {
            return t1;
        }

        int val = t1->val + t2->val;
        TreeNode *node = new TreeNode(val);
        node->left = mergeTrees(t1->left, t2->left);
        node->right = mergeTrees(t1->right, t2->right);
        return node;
    }
};

• Code - Python

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

class Solution(object):
    def mergeTrees(self, t1, t2):
        """
        :type t1: TreeNode
        :type t2: TreeNode
        :rtype: TreeNode
        """
        if t1 == None and t2 == None:
            return None
        if t1 == None:
            return t2
        if t2 == None:
            return t1
        val = t1.val + t2.val
        node = TreeNode(val)
        node.left = self.mergeTrees(t1.left, t2.left)
        node.right = self.mergeTrees(t1.right, t2.right)
        return node

• Code - C

/**
 * 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 == 0 && t2 == 0) {
        return 0;
    }
    if (t1 == 0) {
        return t2;
    }
    if (t2 == 0) {
        return t1;
    }
    struct TreeNode *node = (struct TreeNode*) malloc(sizeof(struct TreeNode));
    node -> val = t1->val + t2->val;
    node -> left = mergeTrees(t1->left, t2->left);
    node -> right = mergeTrees(t1->right, t2->right);
    return node;
}

• Time Complexity: O(N), N is number of tree nodes
• Space Complexity: O(N), N is number of tree nodes