#104 Maximum Depth of Binary Tree
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
解题思路:
1.top_down(root, level)
先得到root节点,这题里level算成1(从例题得知)。那么其子叶的level默认加一。
# 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:
def maxDepth(self, root: Optional[TreeNode]) -> int:
if not root: return 0
global ans
ans = -1
def helper(root, level):
if not root: return
global ans
if not root.left and not root.right: ans = max(ans, level)
helper(root.left, level+1)
helper(root.right, level+1)
helper(root,1)
return ans
runtime:
对于每个节点,其所在的level等于左叶的level最大值与右叶level最大值,两者取的最大值再加一。
# 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:
def maxDepth(self, root: Optional[TreeNode]) -> int:
if not root: return 0
l = self.maxDepth(root.left)
r = self.maxDepth(root.right)
return max(l,r) + 1
runtime:
第二种写起来简单,但是跑起来慢好多。
iteration的思路留给第二次系统刷题的时候做。现在主攻recursion的练习。
#101 Symmetric Tree
Given the root
of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
Input: root = [1,2,2,3,4,4,3] Output: true
Example 2:
Input: root = [1,2,2,null,3,null,3] Output: false
解题思路:
因为要从母节点开始比较,所以适合用top down的办法。
params选择root的左叶, l, 和右叶, r, 传给下一次循环判断。
最简单的base case,分别有三种情况:
1. 只有一个母节点,无左叶和右叶。直接返回True
2.有左叶和右叶,需要做三次bool值的判断:
1)两个值是否相等;
2)再call一次helper function,params为左叶的左叶和右叶的右叶
3)第二次call helper function,params为左叶的右叶和右叶的左叶
当三个bool值都为True的时候,才可以返回True。反之,则False
3. 只有左叶或右叶,直接返回False
最后返回 helper(root.left,root.right)的bool值。
# 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:
def isSymmetric(self, root: Optional[TreeNode]) -> bool:
# check if the values of left pointer and right pointer are the same
def helper(l, r):
# if both nodes are None, return True
if not l and not r:
return True
# if both nodes exist, compare their values
elif l and r:
# True if they are the same
return ((l.val == r.val) and helper(l.left, r.right) and helper(l.right, r.left))
# one exists, the other does not
else:
return False
return helper(root.left, root.right)
runtime:
iterative的思路还是留给第二次系统刷题,当复习。