[LeetCode] 116. Populating Next Right Pointers in Each Node Java

题目：

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

---

题意及分析：给出一棵完美二叉树，即每个非叶子结点都有两个子节点，要求将树的每层用next指针连接起来，每层的最后一个点的next指向null。因为这道题要求使用常数的空间复杂度，所以不能用宽度遍历。这道题使用两个变量，一个pre保存当前层的第一个点，一个cur保存当前遍历到的点。那么对于cur有两种情况，cur是最右的点，那么pre=pre.next向下移动一层；否则对于cur的左子节点的next为cur的右子节点，对于cur的右子节点有成为cur.next的左子节点。

代码：

/**
 * 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;
        TreeLinkNode pre = root;
        TreeLinkNode cur = null;
        while(pre.left!=null){
            cur = pre;
            while(cur!=null){
                cur.left.next = cur.right;
                if(cur.next!=null)
                    cur.right.next = cur.next.left;
                cur = cur.next;
            }
            pre = pre.left;
        }
    }
}