标签：hat wing return none and node tree not ons
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
# 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 buildTree(self, preorder, inorder): """ :type preorder: List[int] :type inorder: List[int] :rtype: TreeNode """ if inorder: ind = inorder.index(preorder.pop(0)) root = TreeNode(inorder[ind]) root.left = self.buildTree(preorder, inorder[0:ind]) root.right = self.buildTree(preorder, inorder[ind+1:]) return root
105. Construct Binary Tree from Preorder and Inorder Traversal
标签：hat wing return none and node tree not ons
原文地址：https://www.cnblogs.com/boluo007/p/12388786.html
