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@@ -1,4 +1,157 @@
+<!--
+ * Author: Jennifer Echenim
+ * GitHub: https://github.com/jennievon
+ * Description: This is a solution to the problem "Get DOM Tree Height", using TypeScript and 3 different approaches - recursive, iterative BFS and iterative DFS
+-->
 
-There is no solution yet.
 
-Would you like to [contribute to the solution](https://github.com/BFEdev/BFE.dev-solutions/blob/main/problem/get-dom-tree-height_en.md)? [Contribute guideline](https://github.com/BFEdev/BFE.dev-solutions#how-to-contribute)
+
+## Recursive solution 
+Recursion involves breaking the problem down into smaller subproblems until we get to a base case. The base case is the simplest possible case that we can solve. Once we solve the base case, we can combine the solutions to the subproblems to solve the original problem.
+
+In this case, the base case is when the tree has no children. The solution to the base case is 1. The solution to the original problem is the max height of the children + 1.
+
+This solution is not as efficient as the iterative solution because it has to traverse the DOM tree multiple times.
+
+Solution:
+1. If the tree is empty, return 0
+2. If the tree has no children, return 1 - base case
+3. If the tree has children, recursively call the function on each child - recursive case
+4. Return the maximum height of the children + 1
+
+Code:
+```javascript
+
+/**
+ * @param {HTMLElement | null} tree
+ * @return {number}
+ */
+
+function getHeight(tree: HTMLElement | null): number {
+  if (!tree) {
+    return 0;
+  }
+
+  // base case
+  if (tree.children.length === 0) {
+    return 1;
+  }
+
+  // recursive case
+  let height = 0;
+  const children = tree.children;
+  for (let i = 0; i < children.length; i++) {
+    height = Math.max(height, getHeight(children[i] as HTMLElement));
+  }
+  return height + 1;
+}
+```
+
+- The **time complexity** of this solution is O(n), where n is the number of nodes because we have to traverse the DOM tree once
+
+- The **space complexity** of this solution is O(n) because we have to store the height of each node in the DOM tree
+
+
+
+
+## Iterative Solution 
+### Using BFS
+BFS involves traversing a tree level by level. We use a queue to keep track of the nodes that need to be visited.
+In this case, we are using a queue to keep track of the nodes to be visited. We will also keep track of the height of the tree as we progress and the the current level.
+
+This solution is more efficient than the recursive solution because it only has to traverse the DOM tree once
+
+Solution:
+1. If the tree is empty, return 0
+2. a queue is initialised with the root node
+3. While this queue is not empty, we increment the height and remove the first element from the queue (FIFO)
+4. Add the children of the removed element to the queue
+5. Return the height
+
+Code:
+```javascript
+function getHeightIterativeBFS(tree: HTMLElement | null): number {
+  if (!tree) {
+    return 0;
+  }
+
+  let height = 0;
+  // use bfs to traverse the tree and keep track of the height
+  const bfsqueue = [tree];
+
+  // while the queue is not empty
+  while (bfsqueue.length) {
+
+    // get the size of the current level
+    const currentLevel = bfsqueue.length;
+
+    for (let i = 0; i < currentLevel; i++) {
+
+      // remove the first element from the queue and add its children to the queue
+      const node = bfsqueue.shift() as HTMLElement;
+
+      // add the children to the queue
+      for (let j = 0; j < node.children.length; j++) {
+        bfsqueue.push(node.children[j] as HTMLElement);
+      }
+    }
+
+    // increment the height after each level is processed
+    height++;
+  }
+  return height;
+}
+```
+
+- The **time complexity** of this solution is O(n) because we have to traverse the DOM tree once
+- The **space complexity** of this solution is O(n) because we have to store the height of each node in the DOM tree
+
+
+
+
+### Using DFS
+DFS involves traversing the tree from the "root to the leaves". We use a stack to keep track of the nodes to be visited.
+In this case, we use a stack to keep track of the nodes to be visited. Just like the BFS approach, we keep track of the height of the tree as we progress and the the current level.
+
+This solution is also more efficient than the recursive solution because it only has to traverse the DOM tree once
+
+Solution:
+1. If the tree is empty, return 0
+2. Initialize a stack with the root node and its level
+3. While the stack is not empty, "pop" the first element from the stack (LIFO)
+4. Update the height to be the max of the current height and the level of the removed element
+5. Add the children of the removed element to the stack
+6. Return the height
+
+Code:
+```javascript
+function getHeightIterativeDFS(tree: HTMLElement | null): number {
+  if (!tree) {
+    return 0;
+  }
+
+  let height = 0;
+  // use dfs to traverse the tree and keep track of the height
+  // use a stack to keep track of the nodes
+  // the stack will contain the node and its level
+  const dfsstack = [[tree, 1]];
+
+  // while the stack is not empty
+  while (dfsstack.length) {
+    // remove the first element from the stack and add its children to the stack
+    const [node, level] = dfsstack.pop() as [HTMLElement, number];
+    height = Math.max(height, level);
+
+    // add the children to the stack
+    for (let j = 0; j < node.children.length; j++) {
+      dfsstack.push([node.children[j] as HTMLElement, level + 1]);
+    }
+  }
+
+  return height;
+}
+```
+
+- The **time complexity** of this solution is O(n) because we have to traverse the DOM tree once
+
+- The **space complexity** of this solution is O(n) because we have to store the height of each node in the DOM tree