Leetcode 617. Merge Two Binary Trees

617. Merge Two Binary Trees

Description

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.

Input:     Tree 1                     Tree 2                            1                         2                                      / \                       / \                                    3   2                     1   3                               /                           \   \                            5                             4   7                  Output: Merged tree:         3        / \       4   5      / \   \      5   4   7

Solution

//题目的意思是给出两个二叉树，将二叉树每个节点的数值相加合并为一个新的二叉树//null视为数值为0，代码较容易理解，如下/* Definition for a binary tree node. struct TreeNode {      int val;      struct TreeNode *left;      struct TreeNode *right;  }; */struct TreeNode* mergeTrees(struct TreeNode* t1, struct TreeNode* t2) {   struct TreeNode *t3 = (struct TreeNode *)malloc(sizeof(struct TreeNode));   if (!t1) //if t1 doesn't exist,take t2 as returned value       return t2;   if (!t2) //if t2 doesn't exist,take t1 as returned value       return t1;   t3->val = t1->val + t2->val;// add two nodes' value   t3->left = mergeTrees(t1->left,t2->left);   t3->right = mergeTrees(t1->right,t2->right);   //create subtrees by recursion   return t3; }