LeetCode 617. Merge Two Binary Trees (C++)

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

思路：从两个树的root开始同时开始做前序遍历，为两树的对应节点做处理，（不引用新的节点）最终返回第一棵树的根节点。

如果两棵树的对应节点都有值，将两值相加并赋给第一棵树的对应节点；

如果第一棵树的对应节点不为空，第二棵树对应节点为空，则不操作（返回）；

如果第一棵树对应节点为空，第二棵树对应节点，将第二棵树的对应节点从第二棵树移到第一棵树的对应位置；

如果两棵树的对应节点都为空，则返回。

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