LeetCode 101. Symmetric Tree - Python

LeetCode 101. Symmetric Tree - Python

 

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

 

Solution : Recursive

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool: 
        if not root: #루트가 없으면 True 반환
            return True
        return self.check(root.left, root.right) #루트의 좌우 값 비교
        
    def check(self, left, right):
        if left is None and right is None: #좌우 둘다 없으면 True
            return True
        if left is None or right is None:  #좌우 중 하나 없으면 False
            return False
        if left.val != right.val:  #좌우 값이 다르면 False
            return False
        a = self.check(left.left, right.right) #좌의 좌 값, 우의 우 값 비교
        b = self.check(left.right, right.left) #좌의 우값, 우의 좌 값 비교
        return a and b 

 

Solution : Iterative

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        if not root:
            return True
        stack = collections.deque([(root.left, root.right)])
        while stack:
            l, r = stack.pop()
            if l is None and r is None:
                continue
            if l is None or r is None:
                return False
            if l.val != r.val:
                return False
            stack.append((l.left, r.right))
            stack.append((l.right, r.left))
        return True

 

 

 

(참고 사이트)