617. Merge Two Binary Trees

最新推荐文章于 2020-05-10 14:18:51 发布

七仔xu

最新推荐文章于 2020-05-10 14:18:51 发布

阅读量227

点赞数

分类专栏：编程

本文链接：https://blog.youkuaiyun.com/qz530308783/article/details/79491093

版权

编程专栏收录该内容

101 篇文章

订阅专栏

本文介绍了一种合并两棵二叉树的算法。当两棵树的节点重叠时，将节点值相加作为新节点值；否则使用非空节点作为新树节点。通过递归方式实现层次遍历并完成合并。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

合并两颗树，需要按照对应层次遍历，将相同未知节点的值相加即可，采用递归完成

/** * 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* mergeTwoNode(TreeNode* t1, TreeNode* t2) { if(t1 ==NULL && t2 == NULL) return NULL; TreeNode *root = new TreeNode(0); if(t1 != NULL && t2 != NULL) root->val = t1->val + t2->val; else if( t1 != NULL && t2 == NULL) root->val = t1->val; else if( t2 != NULL && t1 == NULL) root->val = t2->val; cout<<root->val<<endl; if(t1 != NULL && t2 != NULL) { root->left = mergeTwoNode(t1->left,t2->left); root->right = mergeTwoNode(t1->right,t2->right); }else if(t1 == NULL && t2 != NULL) { root->left = mergeTwoNode(NULL,t2->left); root->right = mergeTwoNode(NULL,t2->right); }else { root->left = mergeTwoNode(t1->left,NULL); root->right = mergeTwoNode(t1->right,NULL); } return root; } TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) { return mergeTwoNode(t1,t2); } };