Populating Next Right Pointers in Each Node

Populating Next Right Pointers in Each Node 

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

 

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

 

For example,
Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

 

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL
这道题提示说要用DFS，但我觉得用BFS好一点，这里我BFS将每一层的所有结点放到一个list中，然后将其放到queue中（这里后面觉得其实没有什么必要，不太想优化了），遍历list中所有元素node，如果node有左子树，将左右结点放到新的list中，遍历完原来的list，在把新的添加到queue中，这里queue作用不太大

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */


import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;


public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null)                                //如果root为空,直接返回
            return;
        else if(null == root.left && null == root.right){    //只有根节点
            root.next = null;
        }
        else{
            Queue<List<TreeLinkNode>> queue = new LinkedList<List<TreeLinkNode>>();        //队列，用于BFS            
            root.next = null;                            //设置root的next
            List<TreeLinkNode> element = new ArrayList<TreeLinkNode>();
            element.add(root.left);                        //左右子节点放到list中，list放到队列中
            element.add(root.right);
            
            queue.add(element);
            //boolean isEnd = false;
            
            while(!queue.isEmpty()){                    //队列不为空
                List<TreeLinkNode> tempList = queue.poll();    //出队
                List<TreeLinkNode> newElement = null;
                
                if(tempList.get(0).left != null){            //有下一层时，在生成新的list
                    newElement = new ArrayList<TreeLinkNode>();
                }
                for(int i = 0 ; i < tempList.size() - 1; i++){    //遍历该层所有节点
                    TreeLinkNode preNode = tempList.get(i);
                    TreeLinkNode behind = tempList.get(i + 1);
                    
                    preNode.next = behind;                        //设置next属性
                    if(preNode.left != null){
                        newElement.add(preNode.left);                //下一层的节点放到list中
                        newElement.add(preNode.right);
                    }                                    
                }
                
                //同一层最后一个结点单独设置
                TreeLinkNode last = tempList.get(tempList.size() - 1);
                last.next = null;
                
                if(null != newElement && null != last.left){
                    newElement.add(last.left);
                    newElement.add(last.right);
                    //将list放到队列中
                    queue.add(newElement);
                }                
            }
        }
    }
}

 在DISCUSS有个写的很PL的

void connect(TreeLinkNode *root) {
    if (root == NULL) return;
    TreeLinkNode *pre = root;
    TreeLinkNode *cur = NULL;
    while(pre->left) {
        cur = pre;
        while(cur) {
            cur->left->next = cur->right;
            if(cur->next) cur->right->next = cur->next->left;
            cur = cur->next;
        }
        pre = pre->left;
    }
}