[Leetcode] 617. Merge Two Binary Trees 解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/78991590

本文介绍了一种将两棵二叉树合并为一棵新二叉树的算法。当两棵树的节点重叠时，新树的节点值为两节点值之和；若仅一棵树有节点，则该节点成为新树相应位置的节点。通过递归实现，适用于计算机科学中的数据结构处理。

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题目：

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.

思路：

练手题目，哈哈。我发现增加一个cloneTree函数之后，可以打败将近90%的提交^_^。

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) {
        TreeNode *t = NULL;
        if (t1 != NULL && t2 != NULL) {
            t = new TreeNode(t1->val + t2->val);
            t->left = mergeTrees(t1->left, t2->left);
            t->right = mergeTrees(t1->right, t2->right);
        }
        else if (t1 != NULL) {  // only t1 is not NULL
            t = cloneTree(t1);
        }
        else if (t2 != NULL) {  // only t2 is not NULL
            t = cloneTree(t2);
        }
        return t;
    }
private:
    TreeNode* cloneTree(TreeNode *t) {
        if (t == NULL) {
            return NULL;
        }
        TreeNode *r = new TreeNode(t->val);
        r->left = cloneTree(t->left);
        r->right = cloneTree(t->right);
        return r;
    }
};