106. 从中序与后序遍历序列构造二叉树

题目描述

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

我的解法——递归（AC）

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def mybuildTree(self, inorder, postorder, l1, r1, l2, r2, dic):
        now = TreeNode(postorder[r2])
        index = dic[postorder[r2]]
        left_count = index - l1
        right_count = r1 - index
        
        
        if(left_count > 0):
            now.left = self.mybuildTree(inorder, postorder, l1, index - 1, l2, l2 + left_count - 1, dic)
        if(right_count > 0):
            now.right = self.mybuildTree(inorder, postorder, index + 1, r1, l2 + left_count, r2 - 1, dic)
        return now
    
    def buildTree(self, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        if(not len(inorder) or not len(postorder)):
            return None
        dic = {}
        for i in range(len(inorder)):
            dic[inorder[i]] = i
        
        return self.mybuildTree(inorder, postorder, 0, len(inorder) - 1, 0, len(postorder) - 1, dic)