题目描述
给定一棵二叉树，设计一个算法，创建含有某一深度上所有节点的链表（比如，若一棵树的深度为 D，则会创建出 D 个链表）。返回一个包含所有深度的链表的数组。
Example1:
输入：[1,2,3,4,5,null,7,8]
1
/ \
2 3
/ \ \
4 5 7 / 8
输出：[[1],[2,3],[4,5,7],[8]]
Note:
- n == points.length
- 2 <= n <= 105
- points[i].length == 2
- 0 <= xi, yi <= 109
解题思路
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
vector<ListNode*> listOfDepth(TreeNode* tree) {
vector<ListNode*> res;
if(tree == nullptr){
return res;
}
queue<TreeNode*> q;
q.push(tree);
while(!q.empty()){
int size = q.size();
ListNode* head = new ListNode(0);
ListNode* p = head;
for(int i = 0; i < size; i++){
TreeNode* cur = q.front();
q.pop();
if(cur -> left){
q.push(cur -> left);
}
if(cur -> right){
q.push(cur -> right);
}
p -> next = new ListNode(cur -> val);
p = p -> next;
}
res.push_back(head -> next);
}
return res;
}
};
