前中序重建二叉树python实现

## 输入某二叉树的前序遍历和中序遍历的结果，
# 请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
# 例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，
# 则重建二叉树并返回。

# -*- coding:utf-8 -*-
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None

class Solution:
# 返回构造的TreeNode根节点
def reConstructBinaryTree(self, pre, tin):
# write code here
if len(pre) == 0 or pre == None :
return None
root = TreeNode(pre[0])
## 获取左子树的前序遍历
child_map = self.getChildRoot(pre, tin)
## 分别获取左子树和右子树的前序遍历和中序遍历
pre_left = child_map['pre_left']
tin_left = child_map['tin_left']
## 递归调用,获取左子树
root.left = self.reConstructBinaryTree(pre_left, tin_left)

pre_right = child_map['pre_right']
tin_right = child_map['tin_right']
root.right = self.reConstructBinaryTree(pre_right, tin_right)
return root

## 找到子树
def getChildRoot(self, pre, tin) :
child = {}
root = pre[0]
pre_left = []
tin_left = []
pre_right = []
tin_right = []
## 中序遍历是否遍历到了根
flag = False
for i in tin :
if i != root and flag != True:
tin_left.append(i)
elif i == root :
flag = True
continue
else :
tin_right.append(i)

for i in pre :
if i in tin_left:
pre_left.append(i)
elif i != root :
pre_right.append(i)
child['pre_left'] = pre_left
child['pre_right'] = pre_right
child['tin_left'] = tin_left
child['tin_right'] = tin_right
return child

## test
pre = [1,2,4,7,3,5,6,8]
tin = [4,7,2,1,5,3,8,6]
s = Solution()
root = s.reConstructBinaryTree(pre, tin)

print(root)