LeetCode--105--从前序与中序遍历序列构造二叉树（python）

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树：

3
/ \
9 20
/ \
15 7

 class Solution:
     def buildTree(self, preorder, inorder):
         """
         :type preorder: List[int]
         :type inorder: List[int]
         :rtype: TreeNode
         """
         def helper(in_left = 0, in_right = len(inorder)):
             nonlocal pre_idx
             # if there is no elements to construct subtrees
             if in_left == in_right:
                 return None

             # pick up pre_idx element as a root
             root_val = preorder[pre_idx]
             root = TreeNode(root_val)

             # root splits inorder list
             # into left and right subtrees
             index = idx_map[root_val]

             # recursion
             pre_idx += 1
             # build left subtree
             root.left = helper(in_left, index)
             # build right subtree
             root.right = helper(index + 1, in_right)
             return root

         # start from first preorder element
         pre_idx = 0
         # build a hashmap value -> its index
         idx_map = {val:idx for idx, val in enumerate(inorder)}
         return helper()

2019-9-20

 class Solution:
     def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
         if len(preorder)==0:
             return None
         if len(preorder)==1:
             return TreeNode(preorder[0])
         root = TreeNode(preorder[0])
         inorderL = inorder[:inorder.index(preorder[0])]
         inorderR = inorder[inorder.index(preorder[0])+1:]
         root.left = self.buildTree(preorder[1:inorder.index(preorder[0])+1],inorderL)
         root.right = self.buildTree(preorder[inorder.index(preorder[0])+1:],inorderR)
         return root

2019-11-20 14:49:13