[LeetCode] 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.

合并二叉树，看到题就想到了递归版本的解法

整个过程分为
1.合并根节点
2.合并左子树
3.合并右子树

TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
    if (t1 == NULL && t2 == NULL)
    {
        return NULL;
    }
    if (t1 == NULL || t2 == NULL)
    {
        return t1 == NULL ? t2 : t1;
    }

    TreeNode* root = new TreeNode(t1->val + t2->val);

    root->left = mergeTrees(t1->left, t2->left);
    root->right = mergeTrees(t1->right, t2->right);

    return root;
}

再看看LeetCode的其他的写法，直接使用传进来的t1指针存储生成的树，生成的树中的节点也是从t1、t2中得到的，省下新建TreeNode的时间

从LeetCode上的提交来看，递归就是这个问题的最优解了，这是我没想到的

TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2)
{
    if (!t1) return t2;
    if (!t2) return t1;

    t1->val += t2->val;
    t1->left = mergeTrees(t1->left, t2->left);
    t1->right = mergeTrees(t1->right, t2->right);

    return t1;
}