117. Populating Next Right Pointers in Each Node II（层序遍历暴力）

 

 

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

  • You may only use constant extra space.

 

For example,
Given the following binary tree,

         1
       /  \
      2    3
     / \    \
    4   5    7

 

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL




层序遍历

 1 class Solution {
 2 public:
 3 
 4     Node* connect(Node* root) {
 5         if(root == NULL) return NULL;
 6         queue<Node*> q;
 7         q.push(root);
 8         while(!q.empty()) {
 9             int level_cnt = q.size();
10             Node* pre_node;
11             Node* cur_node; 
12             for (int i = 0; i < level_cnt; ++i) {
13                 if (i == 0) {
14                     cur_node = q.front();
15                     q.pop();
16                     pre_node = cur_node;
17                 } else {
18                     cur_node = q.front();
19                     q.pop();
20                     pre_node->next = cur_node;
21                     pre_node = cur_node;
22                 }
23                 if(cur_node->left != NULL) q.push(cur_node->left);
24                 if(cur_node->right != NULL) q.push(cur_node->right);
25             }
26             cur_node->next = NULL;
27         }      
28         return root;
29     }
30 };

 






 1 public class Solution {
 2     public void connect(TreeLinkNode root) {
 3         TreeLinkNode head=null;
 4         TreeLinkNode prev =null;
 5         TreeLinkNode cur = root;
 6         
 7         while(cur!=null){
 8             //针对每一层，利用Prev进行next连接
 9             while(cur!=null){
10                 
11                 if(cur.left!=null){
12                     if(prev==null)//prev空，说明cur.left是下一层的头节点。
13                         head = cur.left;
14                     else//不为空，正常连接next指针，
15                         prev.next =cur.left;
16                     prev = cur.left; //移动prev
17                 }
18                 
19                 if(cur.right!=null){
20                  if(prev==null)//prev空，说明cur.right是下一层的头节点。
21                      head = cur.right;
22                 else //不为空，正常连接next指针，
23                      prev.next=cur.right;
24                 prev = cur.right;
25                 }
26                 
27                 cur = cur.next;
28             }
29             //此层结束，准备进入下一层。
30             cur = head;
31             head = null;
32             prev = null;
33         }
34     }
35 }

 

posted @ 2018-01-22 12:05  乐乐章  阅读(140)  评论(0)  编辑  收藏  举报