Populating Next Right Pointers in Each Node II:非完全二叉树构建链表

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.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

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

思路：基于层次遍历，用已构建的链表用于保存下一层的结构信息。

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null) return;
        while(root != null){
            TreeLinkNode head = new TreeLinkNode(0);
            TreeLinkNode temp = head;
            while( root !=null ){
                if(root.left != null){
                    temp.next = root.left;
                    temp = temp.next;
                }
                if(root.right != null){
                    temp.next = root.right;
                    temp = temp.next;
                }
                root = root.next;
            }
            root = head.next;
        }
    }
}