Problem Statement
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:
3
9 20
15 7
Input: root = [3,9,20,null,null,15,7]
Output: 3
Example 2:
Input: root = [1,null,2]
Output: 2
Key Insights
We can
Recursive Solution
# 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
from collections import deque
class Solution:
def maxDepth(self, root: Optional[TreeNode]) -> int:
"""
Time:
* O(N) ~ each node visited once
Space:
* O(N) ~ worst case (one branch)
* O(log N) ~ completely balanced
"""
if not root:
return 0
else:
# For each node, find the height of the
# left and right subtrees
left = self.maxDepth(root.left)
right = self.maxDepth(root.right)
# By adding 1 at each level along the
# longest path, we can find the height
# of the binary tree.
return max(left, right) + 1
Iterative Solution
# 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
from collections import deque
class Solution:
def maxDepth(self, root: Optional[TreeNode]) -> int:
"""
Time:
* O(N) ~ each node visited once
Space:
* O(N) ~ worst case (one branch)
* O(log N) ~ completely balanced
"""
if not root:
return 0
# Add the root to the queue.
queue = deque([(root, 0)])
ans = 0
while queue:
# At each node, increment the height
# and update the global max height
node, height = queue.popleft()
height += 1
ans = max(ans, height)
# Add the left and right node to the
# queue.
if node.left:
queue.append((node.left, height))
if node.right:
queue.append((node.right, height))
return ans