# Question

Formatted question description: https://leetcode.ca/all/116.html

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 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.


Example 2:

Input: root = []
Output: []


Constraints:

• The number of nodes in the tree is in the range [0, 212 - 1].
• -1000 <= Node.val <= 1000

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.

# Algorithm

### DFS

Since it is a complete binary tree, if a node has a left child, then its right child must also exist. Therefore, the next pointer of the left child can directly point to its right child.

To process the right child, we check whether the next pointer of its parent node is empty. If it is not empty, it points to the left child of the node pointed to by its next pointer. If it is empty, it points to NULL.

### BFS

We need to use a queue to assist with the breadth-first search. The nodes in each level are added to the queue in order, and whenever an element is taken from the queue, its next pointer can be pointed to the next node in the queue.

Before starting the traversal of the elements at the beginning of each level, we first count the total number of nodes in the level, and use a for loop to ensure that when the loop ends, the entire level has been traversed.

# Code

• import java.util.LinkedList;
import java.util.Queue;

public class Populating_Next_Right_Pointers_in_Each_Node {

/**
* Definition for binary tree with next pointer.
*     int val;
*     TreeLinkNode(int x) { val = x; }
* }
*/

// just a bfs
public class Solution_iteration {

if (root == null) {
return;
}

q.offer(root);
q.offer(levelEndMarker);

while (!q.isEmpty()) {

if (current == levelEndMarker) {
// push marker again as next level end
if (!q.isEmpty()) { // @note: missed final check again...
q.offer(levelEndMarker);
}

// first node of each level, can not be last of upper level
prev = levelEndMarker; // can be any node which is not in this tree
continue;
}

if (current.left != null) {
q.offer(current.left);
}
if (current.right != null) {
q.offer(current.right);
}

prev.next = current;
prev = current;
}
}
}

public class Solution_recursion {
connect(root, null);
}

if (current == null) {
return;
} else {
current.next = next; // connect
}

connect(current.left, current.right);

if (next != null) {
connect(current.right, next.left);
} else {
/*next == null*/
connect(current.right, null);
}
}
}

}

############

/*
// 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;
}
Deque<Node> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
Node p = null;
for (int n = q.size(); n > 0; --n) {
Node node = q.poll();
if (p != null) {
p.next = node;
}
p = node;
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
}
}
return root;
}
}

• // OJ: https://leetcode.com/problems/populating-next-right-pointers-in-each-node/
// Time: O(N)
// Space: O(N)
class Solution {
public:
Node* connect(Node* root) {
if (!root) return nullptr;
queue<Node*> q{ {root} };
while (q.size()) {
int cnt = q.size();
Node *prev = nullptr;
while (cnt--) {
auto node = q.front();
q.pop();
if (prev) prev->next = node;
prev = node;
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
}
return root;
}
};

• """
# Definition for a Node.
class Node:
def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
self.val = val
self.left = left
self.right = right
self.next = next
"""

'''
>>> d = deque((5,1,9,2))
>>> d.popleft()
5
>>> d.popleft()
1
>>>
>>>
>>> d = deque([5,1,9,2])
>>> d.popleft()
5
>>> d.popleft()
1
'''

from collections import deque

class Solution:
def connect(self, root: "Optional[Node]") -> "Optional[Node]":
if root is None:
return root

# q = deque(root) ===> TypeError: 'Node' object is not iterable
# make it a list [], so it's iterable
q = deque([root])
while q:
p = None
for _ in range(len(q)):
node = q.popleft()
if p:
p.next = node
p = node
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
return root

class Solution:
def connect(self, root: 'Node') -> 'Node':
if not root:
return root

l = [root] # list

while l:
size = len(l)
for i in range(size):
node = l.pop(0) # pop() is last, pop(0) is first of list
if i < size - 1:
node.next = l[0]
if node.left:
l.append(node.left)
if node.right:
l.append(node.right)

return root

# recursion
class Solution:
def connect(self, root: "Optional[Node]") -> "Optional[Node]":
self._connect(root, None)
return root

def _connect(self, current: 'Optional[Node]', next_node: 'Optional[Node]') -> None:
if current is None:
return
else:
current.next = next_node  # connect

self._connect(current.left, current.right)

if next_node is not None:
self._connect(current.right, next_node.left)
else:
self._connect(current.right, None)

• /**
* Definition for a Node.
* type Node struct {
*     Val int
*     Left *Node
*     Right *Node
*     Next *Node
* }
*/

func connect(root *Node) *Node {
if root == nil {
return root
}
q := []*Node{root}
for len(q) > 0 {
var p *Node
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
if p != nil {
p.Next = node
}
p = node
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
}
return root
}

• /**
* Definition for Node.
* class Node {
*     val: number
*     left: Node | null
*     right: Node | null
*     next: Node | null
*     constructor(val?: number, left?: Node, right?: Node, next?: Node) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*         this.next = (next===undefined ? null : next)
*     }
* }
*/

function connect(root: Node | null): Node | null {
if (root == null || root.left == null) {
return root;
}
const { left, right, next } = root;
left.next = right;
if (next != null) {
right.next = next.left;
}
connect(left);
connect(right);
return root;
}