https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
preorder = [3,9,20,15,7] inorder = [9,3,15,20,7]
Return the following binary tree:
3
/ 9 20
/ 15 7
1 # Definition for a binary tree node. 2 # class TreeNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.left = None 6 # self.right = None 7 8 class Solution: 9 def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode: 10 idx = inorder.index( preorder.pop( 0 ) ) 11 12 node = TreeNode( inorder[ idx ] ) 13 14 node.left = buildTree( preorder, inorder[ 0:idx ] ) 15 node.right = buildTree( preorder, inorder[ idx+1: ] ) 16 17 return node
Leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal
原文：https://www.cnblogs.com/pegasus923/p/11203470.html
