[Leetcode]101. Symmetric Tree
📄 Description
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
🔨 My Solution
left
: nodes of the left partsright
: nodes of the right parts 🔵 Rules
1) Conditions for Symmetric Tree
- If the both left and right node is
None
, its Symmetric.- If the one of the left and right node is
None
, its not Symmetric.- If the value is different, its not symmetirc. In other words, the value have to be same to be a symmetric tree.
2) How to check the child nodes : BFS Search
- pop the left element(
pop(0)
) from the left
and right
.- For the left node's child, append the
left.left
first and then left.right
.- For the right node's child, append the
right.right
first and then right.left
.Why is the order you add to queue different?
If you see the above picture, you will understand why.
For the child node(both left and right) popped off from the
left
should be put in order of left->right, and the child node(both left and right) popped off form the right
should be put in order of right->left.💻 My Submission
class Solution:
def isSymmetric(self, root: Optional[TreeNode]) -> bool:
'''
left: left part nodes
right: right part nodes
'''
left=[root.left]
right=[root.right]
while left and right:
l_node=left.pop(0)
r_node=right.pop(0)
# if the both l_node and r_node is None, its symmetric
if not l_node and not r_node:
continue
# if one of the two is None, its not symmetric
if not l_node or not r_node:
return False
# if the value is different, its not symmetirc
if l_node.val!=r_node.val:
return False
# append the left child and right node
left.append(l_node.left)
left.append(l_node.right)
right.append(r_node.right)
right.append(r_node.left)
return True
🎈 Follow up
Could you solve it both recursively and iteratively?
Other Solution (1) Recursion
class Solution:
def isSymmetric(self, root: Optional[TreeNode]) -> bool:
return self.recursion(root.left, root.right)
def recursion(self, l_node,r_node):
if l_node is None and r_node is None:
return True
if l_node is None or r_node is None:
return False
if l_node.val==r_node.val:
return self.recursion(l_node.left, r_node.right) and self.recursion(l_node.right, r_node.left)
else:
return False
Referenceshttps://leetcode.com/problems/symmetric-tree/
https://leetcode.com/problems/symmetric-tree/discuss/33050/Recursively-and-iteratively-solution-in-Python
Reference
この問題について([Leetcode]101. Symmetric Tree), 我々は、より多くの情報をここで見つけました https://velog.io/@limelimejiwon/Leetcode101.-Symmetric-Treeテキストは自由に共有またはコピーできます。ただし、このドキュメントのURLは参考URLとして残しておいてください。
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