leetcode#101 Symmetric Tree

1. 문제

• https://leetcode.com/problems/symmetric-tree/
  

2. 나의 풀이

2.1 in-order 리스트의 결과로 비교하기

# Definition for a binary tree node.
# 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 isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        
        print("root", root)
        
        def in_order(node, result):
            if node and node.left:
                in_order(node.left, result)
            
            if node:
                result.append(node.val)
            
            if node and node.right:
                in_order(node.right, result)
            
            return result
        
        # print("root", root)
        print("in_order", in_order(root, []))
        
        in_order_list = in_order(root, [])
        start_index = 0
        end_index = len(in_order_list) - 1
        while start_index < end_index:
            if in_order_list[start_index] != in_order_list[end_index]:
                return False
            start_index += 1
            end_index -= 1
        return True
            

• input이 [1,2,2,2,null,2]일 때, false여야 함-.-
  
  

2.2 in-order 리스트의 결과로 비교하기 + null인 노드고려하기

# Definition for a binary tree node.
# 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 isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        
        print("root", root)
        
        def in_order(node, result):
            if node and node.left:
                in_order(node.left, result)
            if node and node.left == None and node.right:
                result.append(-1000)
            
            if node:
                result.append(node.val)
            
            if node and node.right:
                in_order(node.right, result)
            if node and node.right == None and node.left:
                result.append(-1000)
                
            
            return result
        
        # print("root", root)
        print("in_order", in_order(root, []))
        
        in_order_list = in_order(root, [])
        start_index = 0
        end_index = len(in_order_list) - 1
        while start_index < end_index:
            if in_order_list[start_index] != in_order_list[end_index]:
                return False
            start_index += 1
            end_index -= 1
        return True

• in_order가 대칭이어여도, tree가 identical하지 않은 케이스 발생
  

2-4. 왼쪽노드의 pre-order와 오른쪽 노드의 post-order의 reverse 값이 같으면 대칭임을 발견!

# Definition for a binary tree node.
# 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 isSymmetric(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        def pre_order(node, result):
            if node:
                result.append(node.val)
                
            if node and node.left:
                in_order(node.left, result)
            if node and node.left == None and node.right:
                result.append(-1000)
            
            if node and node.right:
                in_order(node.right, result)
            if node and node.right == None and node.left:
                result.append(-1000)
            
            return result
                
        def in_order(node, result):
            if node and node.left:
                in_order(node.left, result)
            if node and node.left == None and node.right:
                result.append(-1000)
            
            if node:
                result.append(node.val)
            
            if node and node.right:
                in_order(node.right, result)
            if node and node.right == None and node.left:
                result.append(-1000)
                
            return result
        
        def post_order(node, result):
            if node and node.left:
                in_order(node.left, result)
            if node and node.left == None and node.right:
                result.append(-1000)
            
            if node and node.right:
                in_order(node.right, result)
            if node and node.right == None and node.left:
                result.append(-1000)
            
            if node:
                result.append(node.val)
                
            return result
        
        left = pre_order(root.left, [])
        right = post_order(root.right, [])
        right.reverse()
        
        # print(left, right)
        
        if left == right:
            return True
        return False

3. 남의 풀이