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leetcode-104 Maximum Depth of Binary Tree
- Authors

- Name
- Gene Zhang
[104] Maximum Depth of Binary Tree
Key Concept: Tree Depth Calculation - Use recursion (DFS) or level-order traversal (BFS) to find the maximum depth. Depth is 1 + max depth of subtrees.
DFS approach: Recursive bottom-up calculation. BFS approach: Count levels during level-order traversal.
# Given the root of a binary tree, return its maximum depth.
# A binary tree's maximum depth is the number of nodes along the longest path
# from the root node down to the farthest leaf node.
#
# Example 1:
# Input: root = [3,9,20,null,null,15,7]
# Output: 3
#
# Example 2:
# Input: root = [1,null,2]
# Output: 2
#
# Constraints:
# The number of nodes in the tree is in the range [0, 10^4].
# -100 <= Node.val <= 100
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
# Approach 1: DFS Recursive (Most Concise)
def maxDepth(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))
# Approach 2: BFS Level-Order Traversal
def maxDepth2(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
from collections import deque
queue = deque([root])
depth = 0
while queue:
depth += 1
# Process all nodes at current level
level_size = len(queue)
for _ in range(level_size):
node = queue.popleft()
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return depth
# Approach 3: DFS Iterative with Stack
def maxDepth3(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
stack = [(root, 1)]
max_depth = 0
while stack:
node, depth = stack.pop()
max_depth = max(max_depth, depth)
if node.left:
stack.append((node.left, depth + 1))
if node.right:
stack.append((node.right, depth + 1))
return max_depth
# All approaches: Time O(n), Space O(h) where h is height