判断两棵树是否相同(结构相同，对应的数值相同)
link
显然，这道题可以用DFS
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
|
* 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 (TreeNode* p, TreeNode* q) {
if(p == NULL && q == NULL) return true;
if(p == NULL || q == NULL) return false;
return (p->val==q->val) && isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
}
};
|
这道题第二种解法可以用宽度优先搜索(BFS),此处用栈代替队列
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
|
stack<TreeNode*> stack_p;
stack<TreeNode*> stack_q;
if(p) stack_p.push(p);
if(q) stack_q.push(q);
while(!stack_p.empty() && !stack_q.empty()){
TreeNode* cur_p=stack_p.top();
TreeNode* cur_q=stack_q.top();
stack_p.pop();
stack_q.pop();
if(cur_p->val!=cur_q->val) return false;
if(cur_p->left) stack_p.push(cur_p->left);
if(cur_q->left) stack_q.push(cur_q->left);
if(stack_p.size() != stack_q.size()) return false;
if(cur_p->right) stack_p.push(cur_p->right);
if(cur_q->right) stack_q.push(cur_q->right);
if(stack_p.size() != stack_q.size()) return false;
}
return stack_p.size() == stack_q.size();
|
近期评论