4.3. list of depths

04. Trees and graphs/4.3. list_of_depths.md

@@ -0,0 +1,67 @@
+# 4.3. List of depths
+
+> Given a binary tree, create a linked list of all the nodes at each depth. If you have a tree with depth D, create D linked lists
+
+Create a breadth-first search and at each level create a linked list and append. I don't know how to create a breadth-first algorithm.
+
+## Hints
+
+* A hash table or array mapping from level numbers to nodes at that level might also be helpful
+* First come up with an algorithm which does both depth-first search and breadth-first search
+
+## Solution
+
+We can transverse the graph any way we want, as long as we know which level we're on. We can slightly modify the pre-order traversal algorithm, where we pass in `level+1` to the next recursive call.
+
+```java
+public static void createLevelLinkedList(TreeNode root, ArrayList<LinkedList<TreeNode>> lists, int level) {
+    if (root == null) return;
+    LinkedList<TreeNode> list = null;
+    if (lists.size() == level) { // Level not contained in list
+        list = new LinkedList<TreeNode>();
+        /* Levels are always traversed in order. So, if this is the first time we've visited level i,
+            * we must have seen levels 0 through i - 1. We can therefore safely add the level at the end. */
+        lists.add(list);  
+    } else {
+        list = lists.get(level);
+    }
+    list.add(root);
+    createLevelLinkedList(root.left, lists, level + 1);
+    createLevelLinkedList(root.right, lists, level + 1);
+}
+```
+
+Alternatively, we can also implement a modification of the breadth-first search.
+
+```java
+public static ArrayList<LinkedList<TreeNode>> createLevelLinkedList(TreeNode root) {
+    ArrayList<LinkedList<TreeNode>> result = new ArrayList<LinkedList<TreeNode>>();
+
+    /* "Visit" the root */
+    LinkedList<TreeNode> current = new LinkedList<TreeNode>();
+    if (root != null) {
+        current.add(root);
+    }
+
+    while (current.size() > 0) {
+        result.add(current); // Add previous level
+        LinkedList<TreeNode> parents = current; // Go to next level
+        current = new LinkedList<TreeNode>();
+        for (TreeNode parent : parents) {
+            /* Visit the children */
+            if (parent.left != null) {
+                current.add(parent.left);
+            }
+            if (parent.right != null) {
+                current.add(parent.right);
+            }
+        }
+    }
+
+    return result;
+}
+```
+
+Both algorithms run in O(N) time, but space efficiency? The first option uses O(logN) recursive calls (in a balanced tree), each of which adds a new level to the stack. The second solution, which is iterative doesn't need this extra space.
+
+Both solutions require returning O(N) data. The extra O(logN) used in the first approach is dwarfed by the data that must be returned. So both approaches are equally efficient.

README.md

@@ -47,6 +47,7 @@ If you can't afford to buy the book, you can find a free pdf [here](https://leon
 
 * Find route between nodes in graph
 * Create minimal binary search tree from array
+* Create a linked list for each depth in the tree
 
 ## Chapter 7 Object-oriented design