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leetcode 104. Maximum Depth of Binary Tree （python）
描述
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
Example 3:
Input: root = []
Output: 0
Example 4:
Input: root = [0]
Output: 1
Note:
The number of nodes in the tree is in the range [0, 10^4].
-100 <= Node.val <= 100
解析
根据题意，就是给出了一棵二叉树的根节点 root ，要求我们返回这棵树的最大深度。题目中给出了二叉树的最大深度的定义，就是沿着从根节点到最远叶节点的最长路径的节点个数。其实碰到二叉树的常见问题，你下意识用递归的思路做基本上大概率是靠谱的。
思路比较简单，这里就是定义一个递归函数，如果该节点为空直接 0 ，如果不为空继续加一去探测其左右两边的子树深度，最后返回的结果就是该二叉树的最大深度。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def maxDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1 if root else 0
运行结果
Runtime: 36 ms, faster than 47.47% of Python online submissions for Maximum Depth of Binary Tree.
Memory Usage: 16.2 MB, less than 32.54% of Python online submissions for Maximum Depth of Binary Tree.
解析
当然了也可以不用递归，直接遍历树的每一层节点，当到达新一层的时候最大深度加一，遍历完所有层，即可得到答案。
解答
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def maxDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
result = 0
stack = [root] if root else []
while stack:
result += 1
tmp = []
for node in stack:
if node.left:
tmp.append(node.left)
if node.right:
tmp.append(node.right)
stack = tmp
return result
运行结果
Runtime: 34 ms, faster than 49.72% of Python online submissions for Maximum Depth of Binary Tree.
Memory Usage: 15.9 MB, less than 74.53% of Python online submissions for Maximum Depth of Binary Tree.
原题链接：leetcode.com/problems/ma…
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