leetcode：Populating Next Right Pointers in Each Node II 【Java】

一、问题描述

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

二、问题分析

利用算法leetcode：Binary Tree Level
Order Traversal 【Java】的解题技巧。

三、算法代码

/**
 * 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;
        }
        List<TreeLinkNode> pre = new ArrayList<TreeLinkNode>();//保存树中pre层树结点
        pre.add(root);
        int preLength = 0;
        List<TreeLinkNode> cur = null;//保存树中当前层树结点
        TreeLinkNode curNode = null;
        while(pre.size() != 0){
            preLength = pre.size();
            cur = new ArrayList<TreeLinkNode>();
            for(int i = 0; i <= preLength - 1; i++){
                curNode = pre.get(i);
                if(i == preLength - 1){ //如果是某一层的最后一个结点，则把它的next域置为null
                    curNode.next = null;
                }else{
                    curNode.next = pre.get(i + 1);
                }
                if(curNode.left != null){
                    cur.add(curNode.left); 
                }
                if(curNode.right != null){
                    cur.add(curNode.right); 
                }
            }
            pre = cur;
        }
    }
}