117 Populating Next Right Pointers in Each Node II

题目链接：https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

题目：

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

解题思路：

这题和 116 Populating Next Right Pointers in Each Node 的思路一模一样。实际上，我把 116 的答案直接复制到本题轻松 Accepted 。

具体再说一下思路：

1. 层次遍历二叉树，把层次遍历所需的层队列换成层链表，即每层一个链表

2. 在遍历每一层的过程中，通过一个临时指针 p，将该层的结点链接到一起

3. 每一层遍历完后，返回该层左数第一个结点的引用，也就是该层链表的头结点

4. 在上一层给出的结点链表下，遍历下一层的结点，遍历的过程中重复上述操作

代码实现：

/**
* 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 oldHead = root;
while(oldHead != null) {
TreeLinkNode p = null;
TreeLinkNode newHead = p; // 二者并不是指向同一个对象的引用
while(oldHead != null) {
if(oldHead.left != null) {
if(p == null) {
p = oldHead.left;
newHead = p; // 二者指向同一个对象的引用
}
else {
p.next = oldHead.left;
p = p.next;
}
}
if(oldHead.right != null) {
if(p == null) {
p = oldHead.right;
newHead = p; // 二者指向同一个对象的引用
}
else {
p.next = oldHead.right;
p = p.next;
}
}
oldHead = oldHead.next;
}
oldHead = newHead;
}
}
}

61 / 61 test cases passed.
Status: Accepted
Runtime: 2 ms