LeetCode-617. Merge Two Binary Trees

最新推荐文章于 2020-09-23 09:28:49 发布

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本文介绍了一种合并两棵二叉树的方法，对于重叠节点，其值相加作为新节点的值；对于不重叠节点，则保留非空节点。通过递归实现此过程。

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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.

题目：合并两颗二叉树，相同的节点加和

思路：使用递归的思想，两个节点都存在，则val加和，有节点不存在时，生成一个新节点即可。依次递归两棵树。

代码：

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
    	TreeNode* temp=(TreeNode*)malloc(sizeof(TreeNode));
		if(t1==NULL&&t2==NULL)
			return NULL;
		else if(t1==NULL&&t2!=NULL){
			temp->val=t2->val;
			temp->left = t2->left;
			temp->right = t2->right;
			return temp;
		}
		else if(t1!=NULL&&t2==NULL){
			temp->val=t1->val;
			temp->left=t1->left;
			temp->right=t1->right;
			return temp;
		}	
		else{
			temp->val=t1->val+t2->val;
			temp->left = mergeTrees(t1->left,t2->left);
			temp->right = mergeTrees(t1->right,t2->right);
		}
		return temp; 
    }
};