[Leetcode 617] Merge Two Binary Trees

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].
• -104 <= Node.val <= 104

C code solution:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* mergeTrees(struct TreeNode* root1, struct TreeNode* root2){
    if(!root1 && !root2)
        return NULL;
    if(!root1)
        return root2;
    if(!root2)
        return root1;
    
    root1->val += root2->val;
    
    root1->left = mergeTrees(root1->left, root2->left);
    root1->right = mergeTrees(root1->right, root2->right);
        
    return root1;
}

Explanation:

將root2 merge到root1，利用遞迴。當遇到root1 and root2皆為空時return NULL，而當root1為空則return root2，反之亦然。否則將root2的值加到root1，然後root1->left去接左邊的遞迴，root1->right則接右邊的遞迴，最後回傳root1。