[leetcode] 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

 

https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/

 

思路：关键是5跟6如何连接，用递归的做法，把当前root的左节点和它所有后继右节点 与 右节点和所有深度的左节点连接。

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

 

第二遍记录：

 

第三遍记录： 利用层序遍历的方法，用null分割每一层。每当一个元素弹出queue的时候，该元素的next指针指向队列头部queue.peek()。

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root==null)
            return;
            
        Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
        queue.add(root);
        queue.add(null);
        
        while(!queue.isEmpty()){
            TreeLinkNode out = queue.remove();
            if(out==null){
                if(!queue.isEmpty()){
                    queue.add(null);
                }
            }else{
                out.next = queue.peek();
                if(out.left!=null)
                    queue.add(out.left);
                if(out.right!=null)
                    queue.add(out.right);
            }           
        }
        
    }
}

 

第三遍： 层序遍历

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null)
            return;
        Queue<TreeLinkNode> queue = new LinkedList<>();
        queue.add(root);
        
        while(!queue.isEmpty()){
            int levelNum = queue.size();
            TreeLinkNode pre = null;
            for(int i=0;i<levelNum;i++){
                TreeLinkNode out = queue.remove();
                if(pre!=null){
                    pre.next = out;
                }
                
                if(out.left!=null){
                    queue.add(out.left);
                }
                if(out.right!=null){
                    queue.add(out.right);
                }
                pre = out;
            }
            
        }
        
    }
}

 

 

 

参考：

http://blog.csdn.net/havenoidea/article/details/12840497

http://blog.csdn.net/fightforyourdream/article/details/14514165