617. Merge Two Binary Trees*
https://leetcode.com/problems/merge-two-binary-trees/
题目描述
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.
C++ 实现 1
总的思路是, 先合并左子树, 再合并右子树, 最后合并根节点. 注意边界条件的考虑:
if (!t1 || !t2) return !t1 ? t2 : t1;
这行代码写的很简洁, 它写繁琐一点就是:
if (!t1 && !t2) return nullptr;
if (t1 && !t2) return t1;
if (!t1 && t2) return t2;
完整代码如下:
/**
* 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) {
if (!t1 || !t2) return !t1 ? t2 : t1;
auto left = mergeTrees(t1->left, t2->left);
auto right = mergeTrees(t1->right, t2->right);
TreeNode *root = new TreeNode(t1->val + t2->val);
root->left = left;
root->right = right;
return root;
}
};