LeetCode 116. Populating Next Right Pointers in Each Node

原题链接在这里：https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

题目：

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:

Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

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.

 

题解：

递归调用，如果root.next不为null, root.right.next 等于 root.next.left, 然后root.left.next指向root.right.

Time Complexity: O(n), 每个点访问不超过两遍. Space: O(logn).

AC Java:

/**
 * 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;
        }
        if(root.right != null && root.next != null){
            root.right.next = root.next.left;
        } 
        if(root.left!=null){
            root.left.next = root.right;
        }
        connect(root.right);
        connect(root.left);
    }
}

Iteration版本

Time Complexity: O(n). Space: O(1).

AC Java:

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

    public Node() {}

    public Node(int _val,Node _left,Node _right,Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/
class Solution {
    public Node connect(Node root) {
        Node cur = root;
        while(cur != null){
            Node dummyHead = new Node(); //记录下一层的假头
            Node it = dummyHead;
            
            while(cur != null){
                if(cur.left != null){
                    it.next = cur.left;
                    it = it.next;
                }
                
                if(cur.right != null){
                    it.next = cur.right;
                    it = it.next;
                }
                
                cur = cur.next;  //cur 更新到下一层假头的next上面
            }
            
            cur = dummyHead.next;
        }
        
        return root;
    }
}

跟上Populating Next Right Pointers in Each Node II, Binary Tree Right Side View.