Description
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Note: A leaf is a node with no children.
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its depth = 3.
Solutions
1. 分层遍历
刚好想到最近在看的二叉树分层打印，于是写了如下的解决办法：
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def maxDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
cnt = 0
queue = [root]
last = next_last = root
while queue:
node = queue.pop()
if node.left:
queue.insert(0, node.left)
next_last = node.left
if node.right:
queue.insert(0, node.right)
next_last = node.right
if last == node:
cnt += 1
last = next_last
return cnt
# Runtime: 24 ms, faster than 96.15% of Python online submissions for Maximum Depth of Binary Tree.
# Memory Usage: 14.2 MB, less than 99.43% of Python online submissions for Maximum Depth of Binary Tree.
2. 递归
更好的方法是递归？但是时间不是很高，可能调用函数花了比较多的时间。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def maxDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
def dfs(node):
if not node:
return
l = r = 1
if node.left:
l = dfs(node.left) + 1
if node.right:
r = dfs(node.right) + 1
return max(l, r)
return dfs(root)
# Runtime: 32 ms, faster than 69.54% of Python online submissions for Maximum Depth of Binary Tree.Memory # Usage: 15.2 MB, less than 5.19% of Python online submissions for Maximum Depth of Binary Tree.