題目大意:合併兩棵二叉樹變成一棵樹,合併時重疊的兩節點值相加作爲新節點,否則不爲NULL的節點作爲新節點
分析:dfs。先合併當前節點:有一方爲空就直接返回另一方(這個分支無需再合併了),兩節點都存在就值相加。然後遞歸合併左右子樹即可。
代碼:
/**
* 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) return t2;
if(!t2) return t1;
t1->val += t2->val;
t1->left = mergeTrees(t1->left,t2->left);
t1->right = mergeTrees(t1->right,t2->right);
return t1;
}
};