Leetcode: Populating Next Right Pointers in Each Node II

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

 1 class Solution {
 2     public Node connect(Node root) {
 3         Node pre = null;
 4         if (root == null) return null;
 5         Queue<Node> queue = new LinkedList<>();
 6         queue.offer(root);
 7         while (!queue.isEmpty()) {
 8             int size = queue.size();
 9             for (int i = 0; i < size; i++) {
10                 Node cur = queue.poll();
11                 if (pre == null) pre = cur;
12                 else {
13                     pre.next = cur;
14                     pre = cur;
15                 }
16                 if (cur.left != null) queue.offer(cur.left);
17                 if (cur.right != null) queue.offer(cur.right);
18             }
19             pre = null;
20         }
21         return root;
22     }
23 }

层次递进法

复杂度

时间 O(N) 空间 O(1)

 1 public class Solution {
 2     
 3     //based on level order traversal
 4     public void connect(TreeLinkNode root) {
 5 
 6         TreeLinkNode head = null; //head of the next level
 7         TreeLinkNode prev = null; //the leading node on the next level
 8         TreeLinkNode cur = root;  //current node of current level
 9 
10         while (cur != null) {
11             
12             while (cur != null) { //iterate on the current level
13                 //left child
14                 if (cur.left != null) {
15                     if (prev != null) {
16                         prev.next = cur.left;
17                     } else {
18                         head = cur.left;
19                     }
20                     prev = cur.left;
21                 }
22                 //right child
23                 if (cur.right != null) {
24                     if (prev != null) {
25                         prev.next = cur.right;
26                     } else {
27                         head = cur.right;
28                     }
29                     prev = cur.right;
30                 }
31                 //move to next node
32                 cur = cur.next;
33             }
34             
35             //move to next level
36             cur = head;
37             head = null;
38             prev = null;
39         }
40         
41     }
42 }

posted @ 2014-09-18 07:01 neverlandly 阅读(402) 评论(0) 收藏举报

刷新页面返回顶部