[LeetCode]117. Populating Next Right Pointers in Each Node II

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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

采用层次遍历的方法，对每一层进行操作，同时在操作该层时记下下一层节点的数目

/** * 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;        int thislevel = 1;        int nextlevel = 0;        LinkedList<TreeLinkNode> q = new LinkedList<TreeLinkNode>();        q.add(root);        while(q.isEmpty() == false) {            while(thislevel > 0) {                TreeLinkNode temp = q.poll();                thislevel--;                if (thislevel == 0) temp.next = null;                else temp.next = q.peek();                if(temp.left != null) {                    nextlevel++;                    q.offer(temp.left);                }                if(temp.right != null) {                    nextlevel++;                    q.offer(temp.right);                }            }            thislevel = nextlevel;            nextlevel = 0;        }    }}

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