Populating Next Right Pointers in Each Node II LeetCode Solution

Problem – Populating Next Right Pointers in Each Node II LeetCode Solution

Given a binary tree

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,null,7]
Output: [1,#,2,3,#,4,5,7,#]
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. 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, 6000].
  • -100 <= Node.val <= 100

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.

Populating Next Right Pointers in Each Node II LeetCode Solution in Java

public void connect(TreeLinkNode root) {
    TreeLinkNode dummyHead = new TreeLinkNode(0);
    TreeLinkNode pre = dummyHead;
    while (root != null) {
	    if (root.left != null) {
		    pre.next = root.left;
		    pre = pre.next;
	    }
	    if (root.right != null) {
		    pre.next = root.right;
		    pre = pre.next;
	    }
	    root = root.next;
	    if (root == null) {
		    pre = dummyHead;
		    root = dummyHead.next;
		    dummyHead.next = null;
	    }
    }
}

Populating Next Right Pointers in Each Node II LeetCode Solution in Python

def connect(self, node):
    tail = dummy = TreeLinkNode(0)
    while node:
        tail.next = node.left
        if tail.next:
            tail = tail.next
        tail.next = node.right
        if tail.next:
            tail = tail.next
        node = node.next
        if not node:
            tail = dummy
            node = dummy.next

Populating Next Right Pointers in Each Node II LeetCode Solution in C++

void connect(TreeLinkNode *root) {
    TreeLinkNode *now, *tail, *head;
    
    now = root;
    head = tail = NULL;
    while(now)
    {
        if (now->left)
            if (tail) tail = tail->next =now->left;
            else head = tail = now->left;
        if (now->right)
            if (tail) tail = tail->next =now->right;
            else head = tail = now->right;
        if(!(now = now->next))
        {
            now = head;
            head = tail=NULL;
        }
    }
}
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