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

 1 class Solution:
 2     def buildTree(self, preorder, inorder):
 3         """
 4         :type preorder: List[int]
 5         :type inorder: List[int]
 6         :rtype: TreeNode
 7         """
 8         def helper(in_left = 0, in_right = len(inorder)):
 9             nonlocal pre_idx
10             # if there is no elements to construct subtrees
11             if in_left == in_right:
12                 return None
13             
14             # pick up pre_idx element as a root
15             root_val = preorder[pre_idx]
16             root = TreeNode(root_val)
17 
18             # root splits inorder list
19             # into left and right subtrees
20             index = idx_map[root_val]
21 
22             # recursion 
23             pre_idx += 1
24             # build left subtree
25             root.left = helper(in_left, index)
26             # build right subtree
27             root.right = helper(index + 1, in_right)
28             return root
29         
30         # start from first preorder element
31         pre_idx = 0
32         # build a hashmap value -> its index
33         idx_map = {val:idx for idx, val in enumerate(inorder)} 
34         return helper()

2019-9-20

 1 class Solution:
 2     def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
 3         if len(preorder)==0:
 4             return None
 5         if len(preorder)==1:
 6             return TreeNode(preorder[0])
 7         root = TreeNode(preorder[0])
 8         inorderL = inorder[:inorder.index(preorder[0])]
 9         inorderR = inorder[inorder.index(preorder[0])+1:]
10         root.left = self.buildTree(preorder[1:inorder.index(preorder[0])+1],inorderL)
11         root.right = self.buildTree(preorder[inorder.index(preorder[0])+1:],inorderR)
12         return root

2019-11-20 14:49:13

来源：https://www.cnblogs.com/NPC-assange/p/11511853.html

标签

中序遍历

前序遍历