[LeetCode] Problem 117 - Populating Next Right Pointers in Each Node II

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Given a binary tree

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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

ZWGpK1.png

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

Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note

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

Code

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public class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
}
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public Node connect(Node root) {
    if (root == null)
        return null;

    Node parent = root;
    Node dummy = new Node(0, null, null, null);
    Node current = dummy;

    while (parent != null) {
        if (parent.left != null) {
            current.next = parent.left;
            current = current.next;
        }

        if (parent.right != null) {
            current.next = parent.right;
            current = current.next;
        }

        parent = parent.next;

        if (parent == null) {
            parent = dummy.next;
            dummy.next = null;
            current = dummy;
        }
    }

    return root;
}