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Solution: Maximum Depth (or Height) of Binary Tree

Problem Statement

Given a root node of the binary tree, return the depth (or height) of a binary tree.

The Depth of the binary tree refers to the number of nodes along the longest path from the root node to the farthest leaf node. If the tree is empty, the depth is 0.

Examples

Example 1

  • Input: root = [1, 2, 3, 4, 5]
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  • Expected Output: 3
  • Explanation: The longest path is 1->2->4 or 1->2->5 with 3 nodes.

Example 2

  • Input: root = [1, null, 2, null, 3]
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  • Expected Output: 3
  • Justification: There's only one path 1->2->3 with 3 nodes.

Example 3

  • Input: root = [1, 2, 3, 4, 7, null, null, null, null, null, 9]
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  • Expected Output: 4
  • Justification: The longest path is 1->2->7->9 with 4 nodes.

Constraints:

  • The number of nodes in the tree is in the range [0, 10<sup>4</sup>].
  • -100 <= Node.val <= 100

Solution

What's the 'Depth'?

Picture a ladder. Each step is a level in our tree. The depth? It's just how many steps there are! If we have a 3-step ladder, our tree has a depth of 3.

To determine the deepest level of a binary tree, we'll use a depth-first search (DFS) approach. Begin at the root node and traverse down each branch, keeping track of the current depth. The depth increases by one each time we move to a child node. If a node is a leaf (no children), compare its depth with the current maximum depth and update if necessary. Recursively apply this process to each node's left and right children. This will cover all paths in the tree, allowing us to find and return the maximum depth encountered.

Step-by-step Algorithm

  1. Recursive Approach: The depth of a binary tree can be computed recursively. The maximum depth of a tree is the maximum of the depths of its left and right subtrees plus one (the root node).

  2. Base Condition: If the node is null, return 0. This ensures that we've reached the end of a branch or the tree is empty.

  3. Recursive Calculation: For each node, compute the depth of its left subtree and the depth of its right subtree. The depth of the current node is 1 + the maximum of these two depths.

  4. Result: For the root node, this recursive approach will provide the maximum depth of the entire tree.

Algorithm Walkthrough

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  1. Starting at the root node 1:

    • The method maxDepth(root) is called with the root node 1.
    • The root node is not null, so we proceed to calculate the depth of its left and right subtrees.

  2. Calculating the depth of the left subtree of node 1:

    • The method maxDepth(root.left) is called with the left child of node 1, which is node 2.

  3. Calculating the depth of the left subtree of node 2:

    • The method maxDepth(root.left.left) is called with the left child of node 2, which is node 4.
    • Node 4 is a leaf node (no left or right children).
    • The method maxDepth(root.left.left.left) is called with null (left child of 4), returning 0.
    • The method maxDepth(root.left.left.right) is called with null (right child of 4), returning 0.
    • The maximum depth from node 4 is 1 + max(0, 0) = 1.

  4. Calculating the depth of the right subtree of node 2:

    • The method maxDepth(root.left.right) is called with the right child of node 2, which is node 5.
    • Node 5 is a leaf node (no left or right children).
    • The maximum depth from node 5 is 1 + max(0, 0) = 1.

  5. Determining the depth of node 2:

    • Node 2 has a left depth of 1 (from node 4) and a right depth of 1 (from node 5).
    • The maximum depth from node 2 is 1 + max(1, 1) = 2.

  6. Calculating the depth of the right subtree of node 1:

    • The method maxDepth(root.right) is called with the right child of node 1, which is node 3.
    • Node 3 is a leaf node (no left or right children).
    • The maximum depth from node 3 is 1 + max(0, 0) = 1.

  7. Determining the depth of the root node 1:

    • Node 1 has a left depth of 2 (from node 2) and a right depth of 1 (from node 3).
    • The maximum depth of the tree rooted at node 1 is 1 + max(2, 1) = 3.

Final Output: The maximum depth of the tree [1, 2, 3, 4, 5] is 3.

Code

Here is the code for this algorithm:

Python3
Python3

Complexity Analysis

Time Complexity

  1. Recursive Calls:

    • The algorithm uses recursion to traverse each node in the binary tree exactly once. For each node, it makes a recursive call for the left child and the right child.
    • If the binary tree has n nodes, there will be n recursive calls in total.
    • Therefore, the time complexity for traversing the entire tree is O(n).

  2. Calculating Maximum Depth:

    • For each node, the algorithm calculates the maximum of the left and right subtree depths, which is a constant time operation O(1).

Combining these steps, the overall time complexity of the algorithm is O(n).

Space Complexity

  1. Recursive Stack Space:

    • The depth of the recursion stack is determined by the height of the tree. In the worst case, if the tree is skewed (i.e., all nodes are in a single line), the height of the tree will be n, resulting in a space complexity of O(n) for the recursion stack.
    • In the best case (a balanced tree), the height of the tree will be O(log n), resulting in a space complexity of O(log n).

  2. Additional Space:

    • The algorithm does not use any additional space other than the recursion stack.