【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：

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;
	}