LeetCode 617. Merge Two Binary Trees (Easy)

题目描述：

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:

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

题目大意：给出两棵二叉树，将其二对应节点值相加，形成一棵新树。将节点值相加，如果某一课树的节点不存在就只用将另一棵树对应节点值作为新树的节点值。

思路：同时遍历两棵树并生成新树。二叉树遍历有很多种方法（BFS、DFS（递归非递归）等），下面给出BFS递归遍历。写的比较冗余，要记得判断当前节点是否存在，分情况讨论。
c++代码：

/** * 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) {        int v1 = 0, v2 = 0;        if (t1 != NULL)            v1 = t1->val;        if (t2 != NULL)            v2 = t2->val;        if (t1 == NULL && t2 == NULL)            return NULL;        TreeNode* anst = new TreeNode(v1 + v2);        if (t1 == NULL && t2 != NULL)            anst->left = mergeTrees(NULL, t2->left);        else if (t1 != NULL && t2 == NULL)            anst->left = mergeTrees(t1->left, NULL);        else if (t1 != NULL && t2 != NULL)            anst->left = mergeTrees(t1->left, t2->left);        if (t1 == NULL && t2 != NULL)            anst->right = mergeTrees(NULL, t2->right);        else if (t1 != NULL && t2 == NULL)            anst->right = mergeTrees(t1->right, NULL);        else if (t1 != NULL && t2 != NULL)            anst->right = mergeTrees(t1->right, t2->right);        return anst;    }};

复杂度：O(n)