Python二叉树的镜像转换实现方法示例

# 方式1：生成新的镜像二叉树
def getMirrorBST(self, root):
  if root == None:
    return
  newTree = treeNode(root.val)
  newTree.right = self.getMirrorBST(root.left)
  newTree.left = self.getMirrorBST(root.right)
  return newTree

# 方式2：改变给定的二叉树为镜像二叉树
def turnToMirror(self, root):
  if root == None:
    return
  root.right, root.left = root.left, root.right
  self.turnToMirror(root.left)
  self.turnToMirror(root.right)
  return root

class Solution:
  # 给定一个二叉树，获得其镜像（轴对称）的镜像二叉树：
  # 方式1：生成新的镜像二叉树
  def getMirrorBST(self, root):
    if root == None:
      return
    newTree = treeNode(root.val)
    newTree.right = self.getMirrorBST(root.left)
    newTree.left = self.getMirrorBST(root.right)
    return newTree
  # 方式2：改变给定的二叉树为镜像二叉树
  def turnToMirror(self, root):
    if root == None:
      return
    root.right, root.left = root.left, root.right
    self.turnToMirror(root.left)
    self.turnToMirror(root.right)
    return root
  # 给定二叉树的前序遍历和中序遍历，获得该二叉树
  def getBSTwithPreTin(self, pre, tin):
    if len(pre)==0 | len(tin)==0:
      return None
    root = treeNode(pre[0])
    for order,item in enumerate(tin):
      if root .val == item:
        root.left = self.getBSTwithPreTin(pre[1:order+1], tin[:order])
        root.right = self.getBSTwithPreTin(pre[order+1:], tin[order+1:])
        return root
class treeNode:
  def __init__(self, x):
    self.left = None
    self.right = None
    self.val = x
if __name__ == '__main__':
  flag = "turnToMirror"
  solution = Solution()
  preorder_seq = [1, 2, 4, 7, 3, 5, 6, 8]
  middleorder_seq = [4, 7, 2, 1, 5, 3, 8, 6]
  treeRoot1 = solution.getBSTwithPreTin(preorder_seq, middleorder_seq)
  if flag == "mirrorBST":
    newRoot = solution.getMirrorBST(treeRoot1)
    print(newRoot)
  if flag == "turnToMirror":
    solution.turnToMirror(treeRoot1)
    print(treeRoot1)