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.
1function TreeNode(val, left, right) {
2 this.val = (val===undefined ? 0 : val)
3 this.left = (left===undefined ? null : left)
4 this.right = (right===undefined ? null : right)
5}
6
7var maxDepth = function(root) {
8 if (!root) return 0;
9 return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
10};In JavaScript, similar strategies apply, using recursive function calls to determine the maximum depth by comparing subtree depths. If the node is null, return 0.
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.
1from collections import deque
2
3class TreeNode:
4 def __init__(self, val=0, left=None, right=None):
5 self.val = val
6 self.left = left
7 self.right = right
8
9def maxDepth(root: TreeNode) -> int:
10 if not root:
11 return 0
12 q = deque([root])
13 depth = 0
14 while q:
15 levelSize = len(q)
16 for _ in range(levelSize):
17 node = q.popleft()
18 if node.left:
19 q.append(node.left)
20 if node.right:
21 q.append(node.right)
22 depth += 1
23 return depthUsing Python's collections.deque for efficient pops from a queue's left side, each level’s nodes are tracked and processed, incrementing depth post each level traverse.