leetcode: 105. 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

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


# DFS
class Solution():
    def buildTree(self, preorder, inorder):
        if not len(preorder):
            return None
        root = TreeNode(preorder[0])
        idx = inorder.index(root.val) # 中序中根节点的位置，左边即为左子树
        root.left = self.buildTree(preorder[1 : idx+1], inorder[ : idx])
        root.right = self.buildTree(preorder[idx+1 : ], inorder[idx+1 : ])
        return root