Convert a binary tree to a doubly linked list (in-order)?

Short answer: public class TreeNode { public int val; public TreeNode left, right; public TreeNode(int x) { val = x; } } TreeNode prev = null; TreeNode head = null; TreeNode ConvertToDLL(TreeNode root) { if (root == null) return null; ConvertToDLL(root.left); if (prev == null) { head = root; // first node becomes head } else { root.left = prev; Follow on: prev.right = root; } prev = root; ConvertToDLL(root.right); return head; }…

Explain a bit more

Explanation: Inorder traversal connects nodes as doubly linked list by linking current with previous node.

Example code

public class TreeNode {
public int val;
public TreeNode left, right;
public TreeNode(int x) { val = x; }
}
TreeNode prev = null;
TreeNode head = null; TreeNode ConvertToDLL(TreeNode root) { if (root == null) return null; ConvertToDLL(root.left); if (prev == null) {
head = root; // first node becomes head } else { root.left = prev; Follow on: prev.right = root;
}
prev = root; ConvertToDLL(root.right); return head;
} Explanation: Inorder traversal connects nodes as doubly linked list by linking current with previous node.

Real-world example (ShopNest)

In coding rounds, state complexity aloud, write a clear ShopNest-flavored example (orders, carts), then handle edge cases (empty list, null, overflow).

Say this in the interview

1. Define — one clear sentence (the short answer above).
2. Example — relate it to a project like ShopNest or your real work.
3. Trade-off — when you would not use it.