Populate Next Right pointers of Tree

Dec 18, 2024 • 5 mins read

Problem Overview

LeetCode problem "Populating Next Right Pointers in Each Node" (Problem number: 116) is a classic problem that involves traversing a perfect binary tree and connecting each node to its next right node. If there is no next right node, the pointer should be set to NULL.

The nature of the problem makes it an interesting exercise in understanding tree data structures and pointer manipulation.

Understanding the Problem

Objective

Given a perfect binary tree (where all leaves are at the same level and every parent has two children), the task is to set the next pointer for each node to point to its next right node. If there is no such node, the next pointer should be set to NULL.

Example

Consider the following perfect binary tree:

        1
      /   \
     2     3
    / \   / \
   4   5 6   7

After connecting the next right pointers, the tree should look like this (where # indicates NULL):

        1 -> #
      /   \
     2  -> 3 -> #
    / \   / \
   4->5->6->7 -> #

Constraints

The tree is a perfect binary tree.
We want to achieve this in place (i.e., using no additional space other than the recursion stack).

Key DSA Concepts and Theory

Binary Tree

A Binary Tree is a tree data structure where each node has at most two children referred to as the left child and the right child.

Perfect Binary Tree

A perfect binary tree is a type of binary tree in which every internal node has exactly two children and all leaf nodes are at the same level. This condition ensures that the tree is completely filled at every level except possibly the last.

Pointers and Nodes

In the context of binary trees, nodes are linked using pointers from parent to child. This problem extends this idea by adding a next pointer that connects nodes at the same level.

Solution Approach

Here's a step-by-step explanation of how to solve the problem. The goal is to connect each node's next pointer to its subsequent right node on the same level.

Recursive Solution

Base Case: If the root node is NULL or a leaf node, return since there's nothing to connect.
Current Level Connection: Connect the left child node to the right child node of the same parent.
Cross Level Connection: If there is a next node available for the current node, connect the right child node of the current node to the left child node of its next neighbor.
Recursive Call: Repeat the process for left and right subtrees.

Here's the code implementation in C++:

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

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    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) return root;

        if (root->left) {
            root->left->next = root->right;
            if (root->next) {
                root->right->next = root->next->left;
            }
        }

        connect(root->left);
        connect(root->right);

        return root;
    }
};

And here is how you would implement it in Java:

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;

        if (root.left != null) {
            root.left.next = root.right;
            if (root.next != null) {
                root.right.next = root.next.left;
            }
        }

        connect(root.left);
        connect(root.right);

        return root;
    }
}

Time and Space Complexity Analysis

Time Complexity: O(N), where N is the number of nodes in the tree. This is because each node is visited exactly once.
Space Complexity: O(log N) due to the recursion stack, which is the height of the tree. For a perfect binary tree, the height is log(N).

Common Mistakes to Avoid

Handling Leaf Nodes: Make sure to return immediately when you encounter a leaf node since there are no children to connect.
Cross-Level Connections: Ensure you check if root->next exists before attempting to connect root->right to root->next->left.

Additional Resources and References

These resources provide fundamental concepts that will enhance your understanding of binary tree manipulations and recursive algorithms. Understanding these principles can vastly improve your ability to solve related problems efficiently.

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