【Leetcode】Populating Next Right Pointers in Each Node II

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

Java:

1.征服代码：http://blog.youkuaiyun.com/linhuanmars/article/details/23510601

public void connect(TreeLinkNode root) {
    if(root == null)
        return;
    TreeLinkNode lastHead = root;
    TreeLinkNode pre = null;
    TreeLinkNode curHead = null;
    while(lastHead!=null)
    {
        TreeLinkNode lastCur = lastHead;
        while(lastCur != null)
        {
            if(lastCur.left!=null)
            {
                if(curHead == null)
                {
                    curHead = lastCur.left;
                    pre = curHead;
                }
                else
                {
                    pre.next = lastCur.left;
                    pre = pre.next;
                }
            }
            if(lastCur.right!=null)
            {
                if(curHead == null)
                {
                    curHead = lastCur.right;
                    pre = curHead;
                }
                else
                {
                    pre.next = lastCur.right;
                    pre = pre.next;
                }
            }                
            lastCur = lastCur.next;

        }
        lastHead = curHead;
        curHead = null;
    }
}

后两个还没来得及看懂
2.http://blog.youkuaiyun.com/muscler/article/details/22907093

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    //dfs 
    public void connect(TreeLinkNode root) {
        if (root == null) return;
        
        //如果右孩子不为空，左孩子的next是右孩子。
        //反之，找root next的至少有一个孩子不为空的节点
        if (root.left != null) {
            if (root.right != null) {
                root.left.next = root.right;
            }
            else {
                TreeLinkNode p = root.next;
                while (p != null && p.left == null && p.right == null)
                    p = p.next;
                if (p != null)
                    root.left.next = p.left == null ? p.right : p.left;
            }
        }
        
        //右孩子的next 根节点至少有一个孩子不为空的next
        if (root.right != null) {
            TreeLinkNode p = root.next;
            while (p != null && p.left == null && p.right == null)
                p = p.next;
            if (p != null)
                root.right.next = p.left == null ? p.right : p.left;
        }
        connect(root.right);    
        connect(root.left);
    }
}

3.http://blog.youkuaiyun.com/perfect8886/article/details/20874913

import java.util.LinkedList;
import java.util.Queue;

public class PopulatingNextRightPointersInEachNodeII {
	public void connect(TreeLinkNode root) {
		if (root == null) {
			return;
		}
		Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
		Queue<TreeLinkNode> nextQueue = new LinkedList<TreeLinkNode>();
		queue.add(root);
		TreeLinkNode left = null;
		while (!queue.isEmpty()) {
			TreeLinkNode node = queue.poll();
			if (left != null) {
				left.next = node;
			}
			left = node;
			if (node.left != null) {
				nextQueue.add(node.left);
			}
			if (node.right != null) {
				nextQueue.add(node.right);
			}
			if (queue.isEmpty()) {
				Queue<TreeLinkNode> temp = queue;
				queue = nextQueue;
				nextQueue = temp;
				left = null;
			}
		}
	}
}