Populating Next Right Pointers in Each Node

Given a binary tree

    struct TreeLinkNode {      TreeLinkNode *left;      TreeLinkNode *right;      TreeLinkNode *next;    }

 

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

 

For example,
Given the following perfect binary tree,

         1       /        2    3     / \  /     4  5  6  7

 

After calling your function, the tree should look like:

         1 -> NULL       /        2 -> 3 -> NULL     / \  /     4->5->6->7 -> NULL 

算法思路：

思路1：dfs，将每个节点的左右孩子关联。其次将每个节点的左孩子的右孩子与右孩子的左孩子相关联。

代码略。

思路2：bfs，逐层处理。同学说我针对分层的处理太非主流了。。。。mark一下，以后再处理。

 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 }