Question

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

116. Populating Next Right Pointers in Each Node

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *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.

Note:
    You may only use constant extra space.
    You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,
         1
       /  \
      2    3
     / \  / \
    4  5  6  7
After calling your function, the tree should look like:
         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

@tag-tree

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.
	 * public class TreeLinkNode {
	 *     int val;
	 *     TreeLinkNode left, right, next;
	 *     TreeLinkNode(int x) { val = x; }
	 * }
	 */

    // just a bfs
    public class Solution_iteration {
        public void connect(TreeLinkNode root) {

            if (root == null) {
                return;
            }

            Queue<TreeLinkNode> q = new LinkedList<>();

            TreeLinkNode levelEndMarker = new TreeLinkNode(0);
            TreeLinkNode prev = new TreeLinkNode(0);

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

            while (!q.isEmpty()) {
                TreeLinkNode current = q.poll();

                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 {
        public void connect(TreeLinkNode root) {
            connect(root, null);
        }

        private void connect(TreeLinkNode current, TreeLinkNode next) {
            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

/**
 * 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;
}

116 - Populating Next Right Pointers in Each Node

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Algorithm

DFS

BFS

Code

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Algorithm

DFS

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