题目链接
注意点
解法
小结

题目链接

Same Tree - LeetCode

注意点

先判断结点是否为空再访问结点的值

解法

解法一：递归，从根结点开始判断，然后递归判断左右子树。

/*** 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:bool isSameTree(TreeNode* p, TreeNode* q) {if(!p  && !q) return true;if((p && !q) || (!p && q) || (p->val != q->val)) return false;return isSameTree(p->left,q->left) && isSameTree(p->right,q->right);}
};

解法二：非递归，和递归思想一样，但是显式的用栈来模拟递归的过程。

class Solution {
public:vector<vector<int> > combine(int n, int k) {vector<vector<int> >ret;vector<vector<int> >vec(1);for(int i = 1; i <= n; i++){int len = vec.size();vector<int> tmp;for(int j = 0; j < len; j++){tmp = vec[j];tmp.push_back(i);if(tmp.size() == k)ret.push_back(tmp);else vec.push_back(tmp);}}return ret;}
};

小结

递归的本质就是利用栈实现的

转载于:https://www.cnblogs.com/multhree/p/10500884.html

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Same Tree - LeetCode

题目链接

注意点

解法

小结

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