leetcode 之Populating Next Right Pointers in Each Node II解题思路

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题目如下：

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

这道题和Populating Next Right Pointers in Each Node 那题的区别就是二叉树不再是满二叉树，而是普通的二叉树。所以在和该题的解题上有一点点区别：如何计算每一层的节点个数。当获取到某个节点时，判断该节点的左右孩子是否存在，存在则在下一层的节点数上加1。其它都一样。

/** * 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 || (root.left == null && root.right == null)) return;        Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();        int levelNum = 0;//用来记录当前节点的下一层节点数        if(root.left != null) {queue.add(root.left); levelNum++;}//当左孩子存在时，下一层的节点数加1        if(root.right != null){queue.add(root.right); levelNum++;}//当右孩子存在时，下一层的节点数加1        while(queue.peek() != null){//当队列不为空时，进入循环            int times = levelNum - 1;//控制更改next指针的个数,即：每一层节点个数-1            levelNum = 0;//为记录当前节点的下一层节点数做准备            TreeLinkNode first = null;            TreeLinkNode second = null;            if(times == 0){                first = queue.remove();                if(first.left != null){queue.add(first.left); levelNum++;}                if(first.right != null){queue.add(first.right); levelNum++;}            }            while(times > 1){                first = queue.remove();                second = queue.element();                first.next = second;                if(first.left != null){queue.add(first.left); levelNum++;}                if(first.right != null){queue.add(first.right); levelNum++;}                times--;            }            if(times == 1){              first = queue.remove();              second = queue.remove();              first.next = second;              if(first.left != null){queue.add(first.left); levelNum++;}              if(first.right != null){queue.add(first.right); levelNum++;}              if(second.left != null){queue.add(second.left); levelNum++;}              if(second.right != null){queue.add(second.right); levelNum++;}            }          }                        }}

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