Leetcode 116. Populating Next Right Pointers in Each Node

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本文介绍了一种在二叉树中填充每个节点的next指针的方法，使得每个节点指向其右侧相邻节点。提供了两种解决方案，一种使用递归，另一种利用队列进行层级遍历。

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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,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

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解法1：

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null || root.left == null) {  
            return;  
        }  
        root.left.next = root.right;    
        if(root.next != null) {  
            root.right.next = root.next.left;  
        }  
        connect(root.left);  
        connect(root.right);   
        return;  
    }
}

解法2：

public class Solution {
    private Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>(); 
    private int count = 0;
    public void connect(TreeLinkNode root) {
        if(root == null) return ;
        q.offer(root);
        
        while(!q.isEmpty()){
            TreeLinkNode node = q.poll();
            if(node.left !=null){
                q.offer(node.left);
                q.offer(node.right);
            }
            for(int i = count;i>0;i--){
            	node.next = q.peek();
                node = q.poll();
                if(node.left !=null){
                	q.offer(node.left);
                	q.offer(node.right);
            	}
        	}
            node.next = null;
            count = 2*count +1;
        }
    }
}