List of Depths

Given a binary tree, design an algorithm which creates a linked list of all the nodes at each depth (e.g., if you have a tree with depth D, you'll have D linked lists).

👉 Link here to the repo to solve the problem

👌 Tips

Try modifying a graph search algorithm to track the depth from the root.

A hash table or array that maps from level number to nodes at that level might also be useful.

You should be able to come up with an algorithm involving both depth-first search and breadth-first search.

👊 Solution 1

This solution uses a recursive approach where I create a list of list to keep track of each level, list.get(0) will be the top layer, list.get(1) will be the second layer from the top, etc.

I do it in a very similar approach as preOrderTraversal (reference in the tree section). Where I visit a node, add that value to the list of lists and move onto the children whilst increasing the counter by one.

public class ListOfDepths {
     List<List<Integer>> solution(BinaryTreeNode head) {
         int counter = 0;
         List<List<Integer>> listOfLists = new ArrayList<>();
         traverseAndAppend(listOfLists, head, counter);
         return listOfLists;
     }

     private void traverseAndAppend(List<List<Integer>> list, BinaryTreeNode node, int counter) {
         if (node != null) {
             addToList(list, counter, node.data);
             traverseAndAppend(list, node.left, counter + 1);
             traverseAndAppend(list, node.right, counter + 1);
         }
     }

     private void addToList(List<List<Integer>> list, int counter, int value) {
         try {
             list.get(counter).add(value);
         } catch (IndexOutOfBoundsException error) {
             list.add(counter, new ArrayList<>());
             list.get(counter).add(value);
         }
     }
}

---

Question borrowed from “Cracking the coding interview”