116.Populating-Next-Right-Pointers-in-Each-Node

116. Populating Next Right Pointers in Each Node

题目地址

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

题目描述

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.

Follow up:

You may only use constant extra space.
Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.

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.

Constraints:

The number of nodes in the given tree is less than 4096.
-1000 <= node.val <= 1000

代码

Approach 1: Level Order Traversal BFS

Complexity Analysis

• Time Complexity: O(N) since we process each node exactly once. Note that processing a node in this context means popping the node from the queue and then establishing the next pointers.

• Space Complexity: O(N). This is a perfect binary tree which means the last level contains N/2 nodes. The space complexity for breadth first traversal is the space occupied by the queue which is dependent upon the maximum number of nodes in particular level. So, in this case, the space complexity would be O(N).

/*
// Definition for a Node.
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;
    }
};
*/
class Solution {
  public Node connect(Node root) {
    if (root == null) return root;

    Queue<Node> Q = new LinkedList<Node>();
    Q.add(root);

    while (Q.size() > 0) {
      int size = Q.size();
      for (int i = 0; i < size; i++) {
        Node node = Q.poll();
        if (i < size - 1) { // prevent selection to the next level
          node.next = Q.peek();
        }

        if (node.left != null) {
          Q.add(node.left);
        }
        if (node.right != null) {
          Q.add(node.right);
        }
      }
    }

    return root;
  }
}

Approach #2 Using previously established next points

2 while loop:

1> leftmost = leftmost.left

​ 2> head = head.next

Time Complexity: O_(_N) & Space Complexity: O(1)

class Solution {
  public Node connect(Node root) {
    if (root == null) return root;

    Node leftmost = root;

    while (leftmost.left != null) {
      Node head = leftmost;

     // node and using the next pointers, establish the 
     // corresponding links for the next level
      while (head != null) {
        head.left.next = head.right;
        if (head.next != null) {
          head.right.next = head.next.left;
        }

        head = head.next;
      }

      leftmost = leftmost.left;
    }

    return root;
  }
}