LeetCode 617. Merge Two Binary Trees合并二叉树 (C++)

题目：

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.

分析：

给定两个二叉树，合并他们，也就是将节点处的值加和。

遍历求解，先检查当前节点是否存在，如果一个节点不存在，则返回另一个节点，如果两个节点都存在则将val加和，我们可以复用其中的一颗树，处理完当前节点后，递归求解子树。

程序：

/**
 * 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) return t2;
        if(t2 == nullptr) return t1;
        t1->val += t2->val;
        t1->left = mergeTrees(t1->left, t2->left);
        t1->right = mergeTrees(t1->right, t2->right);
        return t1;
    }
};

原文地址：https://www.cnblogs.com/silentteller/p/10747717.html