【LeetCode】617. Merge Two Binary Trees
- 2019 年 11 月 8 日
- 筆記
版權聲明：本文為部落客原創文章，遵循 CC 4.0 BY-SA 版權協議，轉載請附上原文出處鏈接和本聲明。
本文鏈接：https://blog.csdn.net/shiliang97/article/details/102181597
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.
來源：力扣（LeetCode） 鏈接：https://leetcode-cn.com/problems/merge-two-binary-trees 著作權歸領扣網路所有。商業轉載請聯繫官方授權，非商業轉載請註明出處。
class Solution { public: TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) { if(t1==NULL&& t2!=NULL) return t2; if(t1 !=NULL&&t2!=NULL){ t1->val=t1->val+t2->val; t1->left = mergeTrees(t1->left,t2->left); t1->right = mergeTrees(t1->right,t2->right); } return t1; } };
這樣更快點
class Solution { public: TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) { if(t1==NULL&& t2!=NULL){ return t2; } if (t2==NULL&& t1!=NULL){ return t1; } if(t1 !=NULL&&t2!=NULL){ int n=t1->val+t2->val; t1->val=n; t1->left = mergeTrees(t1->left,t2->left); t1->right = mergeTrees(t1->right,t2->right); } return t1; } };
