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LeetCode : 116. Populating Next Right Pointers in Each Node

2022-05-23
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LeetCode

116. Populating Next Right Pointers in Each Node

Problem

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: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]

[1] Code (22. 05. 23)

Need to Retry -> 올바른 아이디어를 못떠올렸다.

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

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

        while (!q.isEmpty()) {
            int size = q.size();
            Node prev = new Node();
            for (int i = 0; i < size; i++) {
                Node node = q.poll();
                prev.right = node;
                prev = node;

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

        return root;
    }
}

  • (220523)

    TC를 돌리면 틀리는데

    위 로직이 왜 틀린건지 모르겠다.

Reference Code

Code 1

// Runtime: 0 ms
// Memory Usage: 47 MB
// Ref : https://leetcode.com/submissions/detail/704856906
class Solution {
    public Node connect(Node root) {
        if (root == null) {return null;}
        if (root.left != null) {root.left.next = root.right;}
        if (root.right != null && root.next != null) {root.right.next = root.next.left;}
        connect(root.left);
        connect(root.right);
        return root;
    }
}
  • 문제 Discuss 코드를 참고했다.

Code 2

// Runtime: 4 ms
// Memory Usage: 47.3 MB
// Ref : https://leetcode.com/submissions/detail/704860727
class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return null;
        }
        link(root.left, root.right);
        return root;
    }

    public void link(Node left, Node right) {
        if (left == null && right == null) {
            return;
        }
        left.next = right;
        link(left.left, left.right);
        link(left.right, right.left);
        link(right.left, right.right);
    }
}
  • 문제 Discuss 코드를 참고했다.

Review

  • 이런 유형의 문제를 해결하는 다양한 접근법을 배웠다.

[2] Code (23. 02. 18) (x)

다시 풀 필요는 없으나 정답 아이디어만 훑어보자.

// Runtime: 3 ms
// Memory Usage: 42.2 MB
// Ref : https://leetcode.com/submissions/detail/900201642
class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return null;
        }
        
        Queue<Node> q = new LinkedList<>();
        q.add(root);
        
        while (!q.isEmpty()) {
            int size = q.size();
            
            List<Node> list = new ArrayList<>();
            
            for (int i=0; i<size; i++) {
                Node node = q.poll();
                list.add(node);
                
                if (node.left != null) {
                    q.add(node.left);
                }
                
                if (node.right != null) {
                    q.add(node.right);
                }
            }
            
            int listSize = list.size();
            for (int i=0; i<listSize; i++) {
                if (i == listSize - 1) {
                    list.get(i).next = null;
                } else {
                    list.get(i).next = list.get(i+1);
                }
            }
        }

        return root;
    }
}
  • 1, 2, 3 코드를 살펴보자.

Reference

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