Brooke

一、題目

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

struct Node {  int val;  Node *left;  Node *right;  Node *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.

 

Example 1:

 

 

 

 

 

Input: root = [1,2,3,4,5,6,7] Output: [1,#,2,3,#,4,5,6,7,#] Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Example 2:

Input: root = [] Output: []

 

Constraints:

• The number of nodes in the tree is in the range [0, 212 - 1].
• -1000 <= Node.val <= 1000

 

Follow-up:

• You may only use constant extra space.
• The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.

 

二、程式作法

 /*寫法一*/
/*
// Definition for a Node.
public class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
}
*/

public class Solution
{
    public Node Connect(Node root)
    {
        if (root != null)
            ToConnect(root);

        return root;
    }

    public void ToConnect(Node root)
    {
        if (root.left != null)
            root.left.next = root.right;

        if (root.next != null && root.right != null)
            root.right.next = root.next.left;

        if (root.left != null)
        {
            ToConnect(root.left);
            ToConnect(root.right);
        }
    }
}

/*寫法二*/
/*
public class Solution
{
    public Node Connect(Node root)
    {
        if (root == null || root.left == null)
            return root;
        else
            root.left.next = root.right;

        if (root.next != null && root.right != null)
            root.right.next = root.next.left;

        Connect(root.left);
        Connect(root.right);
        return root;
    }
}
*/

 

三、思路筆記

目的為針對一顆 perfect binary tree 將其 next 指標指向同階層的右節點。

每一節點所要做的事有二，

將左節點的指標指到右節點，再將右節點指標指到原節點指標的左節點。