617. Merge Two Binary Trees 二叉树合并

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.
将给定的两个二叉树合并，对应节点都存在的话将值相加作为合并后的节点值。

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.

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思路
【方法1】递归

/** * 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) 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;    }};

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {        if(t1==null) return t2;        if(t2==null) return t1;        t1.val += t2.val;        t1.left = mergeTrees(t1.left,t2.left);        t1.right = mergeTrees(t1.right,t2.right);        return t1;    }}

【方法2】迭代