【Leetcode】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

思路：
1、用"Populating
Next Right Pointers in Each Node".的算法，仍然有效，只是空间复杂度是O(n)

算法1：

public void connect(TreeLinkNode root) {
List<List<TreeLinkNode>> lists = levelOrder(root);
for (List<TreeLinkNode> list : lists) {
for (int i = 0; i < list.size() - 1; i++) {
list.get(i).next = list.get(i + 1);
}
list.get(list.size() - 1).next = null;
}
}

/**
* 统计每层节点
*/
public List<List<TreeLinkNode>> levelOrder(TreeLinkNode root) {
int height = heightTree(root);
List<List<TreeLinkNode>> lists = new ArrayList<List<TreeLinkNode>>();
for (int i = 1; i <= height; i++) {
List<TreeLinkNode> list = new ArrayList<TreeLinkNode>();
list = kLevelNumber(root, 1, list, i);
lists.add(list);
}
return lists;
}

/***
* kk是目标层数，height是当前遍历结点高度
*/
public List<TreeLinkNode> kLevelNumber(TreeLinkNode p, int height, List<TreeLinkNode> list, int kk) {
if (p != null) {
if (height == kk) {
list.add(p);
}
list = kLevelNumber(p.left, height + 1, list, kk);
list = kLevelNumber(p.right, height + 1, list, kk);
}
return list;
}

public int heightTree(TreeLinkNode p) {
if (p == null)
return 0;
int h1 = heightTree(p.left);
int h2 = heightTree(p.right);
return h1 > h2 ? h1 + 1 : h2 + 1;
}