【Leetcode】617. Merge Two Binary Trees

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本文介绍了一种将两棵二叉树合并为一棵新二叉树的方法，对于重叠节点，其值为两树相应节点值之和；若节点不重叠，则采用非空节点值。通过递归方式实现该算法，并提供了C++代码示例。

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Description:

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:

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.

思路：

本题要求是将两个二叉树合并成一个新的二叉树，对应位置的值相加。那么这里可以利用树的性质进行递归。从根节点开始，对其左右两边分别进行递归，直至所有的点访问完毕，最后返回这棵树。

下面是使用C++的实现过程：

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        if(t1==NULL) return t2;
        if(t2==NULL) return t1;

        TreeNode* t = new TreeNode(t1->val + t2->val);
        t->left = mergeTrees(t1->left,t2->left);
        t->right = mergeTrees(t1->right,t2->right);
        return t;
    }
};