Last updated on October 5th, 2024 at 09:09 pm

Here, We see Populating Next Right Pointers in Each Node II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Populating Next Right Pointers in Each Node II LeetCode Solution

Problem Statement

Given a binary treestruct 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: []

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

class Solution {
public:
    Node* connect(Node* root) {
        Node *currParent = root, *baseChild, *currChild, *nextChild;
        while (currParent) {
            while (currParent->next && !currParent->left && !currParent->right) currParent = currParent->next;
            currChild = baseChild = currParent->left ? currParent->left : currParent->right;
            while (currChild) {
                if (currParent->right && currChild != currParent->right) nextChild = currParent->right;
                else {
                    currParent = currParent->next;
                    while (currParent && !currParent->left && !currParent->right) currParent = currParent->next;
                    nextChild = currParent ? currParent->left ? currParent->left : currParent->right : currParent;
                }
                currChild->next = nextChild;
                currChild = nextChild;
            }
            currParent = baseChild;
        }
        return root;
    }
};

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

class Solution {
    public Node connect(Node root) {
        if (root == null || (root.left == null && root.right == null)) {
            return root;
        }
        if (root.left != null && root.right != null) {
            root.left.next = root.right;
            root.right.next = getNextHasChildrenNode(root);
        }
        if (root.left == null) {
            root.right.next = getNextHasChildrenNode(root);
        }
        if (root.right == null) {
            root.left.next = getNextHasChildrenNode(root);
        }
        root.right = connect(root.right);
        root.left = connect(root.left);
        return root;
    }
    public Node getNextHasChildrenNode(Node root) {
        while (root.next != null) {
            if (root.next.left != null) {
                return root.next.left;
            }
            if (root.next.right != null) {
                return root.next.right;
            }
            root = root.next;
        }
        return null;
    }
}

3. Populating Next Right Pointers in Each Node II Solution JavaScript

var connect = function(root) {
    if (!root) return root;
    let queue = [root];
    let tempQueue = [];
    while(queue.length){
        let curr = queue.splice(0, 1)[0];
        let {left, right} = curr;    
        if (left) tempQueue.push(left);
        if (right) tempQueue.push(right);
        if (queue.length === 0){
            curr.next = null;
            queue = tempQueue;
            tempQueue = [];
        }else{
            curr.next = queue[0];
        }
    }
    return root;
};

4. Populating Next Right Pointers in Each Node II Solution Python

class Solution(object):
    def connect(self, root):
        if not root:
            return None
        q = deque()
        q.append(root)
        dummy=Node(-999)
        while q:
            length=len(q)
            prev=dummy
            for _ in range(length):
                popped=q.popleft()
                if popped.left:
                    q.append(popped.left)
                    prev.next=popped.left
                    prev=prev.next
                if popped.right:
                    q.append(popped.right)
                    prev.next=popped.right
                    prev=prev.next
        return root