[leetcode]106.从中序与后序遍历序列构造二叉树

根据一棵树的中序遍历与后序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]

返回如下的二叉树：

    3
   / 
  9  20
    /  
   15   7

class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
        n = len(inorder)
        m = len(postorder)
        if m != n:
            return None
        mapper = {}
        for index,num in enumerate(inorder):
            mapper[num] = index
        return self.build(postorder,0,n-1,mapper,0,m-1)
    def build(self,postorder,postLeft,postRight,mapper,inLeft,inRight):
        if postLeft>postRight or inLeft>inRight:
            return None
        root_val = postorder[postRight]
        root = TreeNode(root_val,None,None)
        index = mapper.get(root_val)
        root.left = self.build(postorder,postLeft,postRight-inRight+index-1,mapper,inLeft,index-1)
        root.right = self.build(postorder,postRight-inRight+index,postRight-1,mapper,index+1,inRight)
        return root