This approach involves using a recursive function that traverses the tree in a depth-first manner. For each node, calculate the maximum depth of its left and right subtrees, and add 1 for the current node itself. The function returns the maximum of these two values. This provides an elegant and intuitive solution, leveraging the inherent recursive structure of trees.
Time Complexity: O(n) where n is the number of nodes, as each node is visited once.
Space Complexity: O(h) where h is the height of the tree, due to the stack space in recursion.
1class TreeNode:
2 def __init__(self, val=0, left=None, right=None):
3 self.val = val
4 self.left = left
5 self.right = right
6
7def maxDepth(root: TreeNode) -> int:
8 if not root:
9 return 0
10 return max(maxDepth(root.left), maxDepth(root.right)) + 1This Python solution verifies if the tree root exists. If not, it returns 0. Otherwise, it recursively calculates the maximum depth of both subtrees, adding 1 to account for the current node.
This approach involves using a queue to perform a Breadth-First Search (BFS) on the tree. By iterating level by level, we increment the depth counter with each level traversed completely.
Time Complexity: O(n) due to each node being visited once.
Space Complexity: O(n) where n is the maximum number of nodes at any level.
1import java.util.LinkedList;
2import java.util.Queue;
3
4public class TreeNode {
5 int val;
6 TreeNode left;
7 TreeNode right;
8 TreeNode(int x) { val = x; }
9}
10
11class Solution {
12 public int maxDepth(TreeNode root) {
13 if (root == null) return 0;
14 Queue<TreeNode> q = new LinkedList<>();
15 q.offer(root);
16 int depth = 0;
17 while (!q.isEmpty()) {
18 int levelSize = q.size();
19 for (int i = 0; i < levelSize; i++) {
20 TreeNode node = q.poll();
21 if (node.left != null) q.offer(node.left);
22 if (node.right != null) q.offer(node.right);
23 }
24 depth++;
25 }
26 return depth;
27 }
28}This Java solution uses a Queue to explore levels of the tree node by node. offer() inserts children nodes for subsequent processing, while poll() removes nodes, level by level incrementing depth.