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树-leetcode-101


​​0️⃣python数据结构与算法学习路线​​ 学习内容:

  • 基本算法:枚举、排序、搜索、递归、分治、优先搜索、贪心、双指针、动态规划等…
  • 数据结构:字符串(string)、列表(list)、元组(tuple)、字典(dictionary)、集合(set)、数组、队列、栈、树、图、堆等…

题目:

给定一个二叉树,检查它是否是镜像对称的。
例如,二叉树 [1,2,2,3,4,4,3] 是对称的。


解题思路:

递归方法:
判断根是不是为空
不为空将左右子树输入函数
判断左子树的左子树和右子树的右子树,左子树的右子树和右子树的左子树是否相同

算法实现:

# 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: TreeNode) -> bool:
if root is None:
return True
def childtree(left,right):
if not (left or right):
return True
elif not (left and right):
return False
elif left.val != right.val:
return False
return childtree(left.left,right.right) and childtree(left.right,right.left)
return childtree(root.left,root.right)

出现问题:

本地实现

class TreeNode:
def __init__(self, value):
self.left = None
self.val = value
self.right = None

def insertLeft(self, value):
self.left = TreeNode(value)
return self.left

def insertRight(self, value):
self.right = TreeNode(value)
return self.right

def show(self):
print(self.val)

def isSymmetric( root: TreeNode)->bool:
if root is None:
return True

def childtree(left,right):
if not (left or right):
return True
elif not (left and right):
return False
elif left.val != right.val:
return False
return childtree(left.left,right.right) and childtree(left.right,right.left)
return childtree(root.left,root.right)
if __name__ == '__main__':
Root = TreeNode('1')
Root.show()
A = Root.insertLeft('2')
A.insertLeft('3')
A.insertRight('4')
B = Root.insertRight('2')
B.insertRight('3')
B.insertLeft('4')

print(isSymmetric(root=Root))


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