LeetCode 101. Symmetric Tree
https://leetcode.com/problems/symmetric-tree/
Description
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
Solution
Solution 1
40 ms AC.
如果root是对称的，那么他的左右子树对称。递归验证对称性
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# judge whether root1 and root2 are symmetric
def judge(self, root1, root2):
if not root1 and not root2: # all empty
return True
elif (not root1) or (not root2): # one is empty, the other is not
return False
elif root1.val != root2.val:
return False
symmetric1, symmetric2 = False, False
if not root1.left and not root2.right: # these two are empty
symmetric1 = True
elif root1.left and root2.right:
symmetric1 = self.judge(root1.left, root2.right)
if not root1.right and not root2.left: # these two are empty
symmetric2 = True
elif root1.right and root2.left:
symmetric2 = self.judge(root1.right, root2.left)
return symmetric1 and symmetric2
def isSymmetric(self, root: 'TreeNode') -> 'bool':
if not root:
return True
# Recursively judge whether left of root and right of root are symmetric
return self.judge(root.left, root.right)
Solution 2
44 ms AC.
对称的必要条件：
- 中序遍历得到的序列应该是左右对称的 (191/193 passed)
- 中序遍历和逆中序遍历得到的序列应该是相同的 (192/193 passed)
用条件2做，可以达到 192/193 passed。只有一个特殊样例过不了：[5,4,1,null,1,null,4,2,null,2,null]
因为这棵树的中序和逆中序序列是完全一样的。
5
/ \
4 1
/ \ \
1 4
/ /
2 2
特殊判断一下root的左右子树是否相等，可以AC。此做法不推荐，投机取巧之嫌，而且比法1麻烦。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
nodelist = []
# 中序遍历
# root.left, root, root.right
def inorder(self, root):
if not root:
return
isLeaf = False
if not root.left and not root.right:
isLeaf = True
if root.left:
self.inorder(root.left)
elif not isLeaf:
self.nodelist.append(None)
self.nodelist.append(root.val)
if root.right:
self.inorder(root.right)
elif not isLeaf:
self.nodelist.append(None)
# 逆中序遍历
# root.right, root, root.left
def reverseInorder(self, root):
if not root:
return
isLeaf = False
if not root.left and not root.right:
isLeaf = True
if root.right:
self.reverseInorder(root.right)
elif not isLeaf:
self.nodelist.append(None)
self.nodelist.append(root.val)
if root.left:
self.reverseInorder(root.left)
elif not isLeaf:
self.nodelist.append(None)
def judge(self, lst1, lst2):
len1, len2 = len(lst1), len(lst2)
if len1 != len2:
return False
for i in range(len1):
if lst1[i] != lst2[i]:
return False
return True
def isSymmetric(self, root: 'TreeNode') -> 'bool':
self.nodelist = []
self.inorder(root)
lst1 = self.nodelist.copy()
# print(self.nodelist)
self.nodelist = []
self.reverseInorder(root)
lst2 = self.nodelist.copy()
# print(self.nodelist)
if self.judge(lst1, lst2):
# special judgement
if root and root.left and root.right:
if root.left.val != root.right.val:
return False
return True
return False