Python程序:检查给定树是否是对称树


假设我们有一棵二叉树。我们需要检查这棵树是否是对称树。如果一棵树与其镜像相同,则称其为对称树。从这两棵树来看,第一棵是对称的,但第二棵不是。

为了解决这个问题,我们将遵循以下步骤。

  • 我们将递归地调用以下步骤。该函数将是solve(root, root)

  • 如果节点1和节点2为空,则返回true

  • 如果节点1或节点2为空,则返回false

  • 当node1.val = node2.val 且 solve(node1.left, node2.right) 且 solve(node1.right, node2.left) 时返回true

让我们来看下面的实现,以便更好地理解——

示例

 在线演示

class TreeNode:
   def __init__(self, data, left = None, right = None):
      self.data = data
      self.left = left
      self.right = right
class Solution(object):
   def isSymmetric(self, root):
      return self.solve(root,root)
   def solve(self,node1,node2):
      if not node1 and not node2:
         return True
      if not node1 or not node2:
         return False
      return node1.data == node2.data and
self.solve(node1.left,node2.right) and
self.solve(node1.right,node2.left)
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(2)
root.left.left = TreeNode(3)
root.left.right = TreeNode(4)
root.right.left = TreeNode(4)
root.right.right = TreeNode(3)
ob1 = Solution()
print(ob1.isSymmetric(root))

输入

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(2)
root.left.left = TreeNode(3)
root.left.right = TreeNode(4)
root.right.left = TreeNode(4)
root.right.right = TreeNode(3)

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输出

True

更新于:2020年10月21日

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