[105] Construct Binary Tree from Preorder and Inorder Traversal

https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/

• algorithms
• Medium (37.84%)
• Source Code: 105.construct-binary-tree-from-preorder-and-inorder-traversal.py
• Total Accepted: 213.4K
• Total Submissions: 528.2K
• Testcase Example: '[3,9,20,15,7]\n[9,3,15,20,7]'

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 not preorder or not inorder:
            return None
        node = TreeNode(preorder[0])
        idx = inorder.index(preorder[0])
        preorder.pop(0)
        node.left = self.buildTree(preorder, inorder[0:idx])
        node.right = self.buildTree(preorder, inorder[idx+1:])
        return node