算法分析与设计——LeetCode Problem.617 Merge Two Binary Trees

问题详情

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.

数据结构为：

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
}

问题分析及思路

这道题是要讲两个二叉树合并，新的二叉树的每个位置的值为原来两个二叉树对应位置的值之和，运用递归解决即可。

算法复杂度与二叉树的最高层数有关。

具体代码

/**
* 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) {
if (!t1) return t2;
if (!t2) return t1;

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