特定深度节点链表

题目链接

 

 解题思路：BFS+链表

水题:基本BFS

用BFS一层一层遍历一遍，把同一层次的数存入同一个链表链表，最后输出结果即可

/**
 * 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*>vt;
        queue<TreeNode*>q;
        if(tree==NULL){
            return vt;
        }
        q.push(tree);
        int len;
        while(!q.empty()){
            ListNode* head=new ListNode();
            ListNode* tail=new ListNode();
            len=q.size();
            for(int i=1;i<=len;i++){
                TreeNode* newNode=q.front();
                q.pop();
                if(i==1){
                    head->val=newNode->val;
                    tail = head;
                }
                else{
                    tail->next=new ListNode(newNode->val);
                    tail=tail->next;
                }
                if(newNode->left){
                    q.push(newNode->left);
                }
                if(newNode->right){
                    q.push(newNode->right);
                }
            }
            vt.push_back(head);
        }
        return vt;
    }
};