[leetcode]Populating Next Right Pointers in Each Node II

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

算法思路：

BFS，空间复杂度不是常数【MARK】

偷懒了，跟[leetcode]Populating Next Right Pointers in Each Node一样处理了。

代码如下：

 1 public class Solution {
 2     public void connect(TreeLinkNode root) {
 3         if(root == null) return;
 4         Queue<TreeLinkNode> q = new ArrayDeque<TreeLinkNode>();
 5         q.offer(root);
 6         root.next = null;
 7         Queue<TreeLinkNode> copy = new ArrayDeque<TreeLinkNode>();
 8         while(!q.isEmpty()){
 9             TreeLinkNode tem = q.poll();
10             tem.next = q.peek();
11             if(tem.left != null) copy.offer(tem.left);
12             if(tem.right != null) copy.offer(tem.right);
13             if(q.isEmpty()){
14                 for(TreeLinkNode node : copy){
15                     q.offer(node);
16                 }
17                 copy.clear();
18             }
19         }
20     }
21 }

第二遍：

用null来标记某一行的结束。

代码如下：

 1 /**
 2  * Definition for binary tree with next pointer.
 3  * public class TreeLinkNode {
 4  *     int val;
 5  *     TreeLinkNode left, right, next;
 6  *     TreeLinkNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     public void connect(TreeLinkNode root) {
11         if(root == null || (root.left == null && root.right == null)) return;
12         Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
13         q.offer(root);
14         q.offer(null);
15         while(!q.isEmpty()){
16             TreeLinkNode tem = q.poll();
17             if(tem != null){
18                 tem.next = q.peek();
19                 if(tem.left != null)  q.offer(tem.left);
20                 if(tem.right != null) q.offer(tem.right);
21             }else{
22                 if(!q.isEmpty()) q.offer(null);
23             }
24         }
25     }
26 }