# Question

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


# Solution

Find the corresponding part for left tree and right tree, and build the tree recursively.

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

class Solution:
def buildTree(self, preorder, inorder):
"""
:type preorder: List[int]
:type inorder: List[int]
:rtype: TreeNode
"""
if len(preorder) == 0:
return None
if len(preorder) == 1:
return TreeNode(preorder[0])

root = preorder[0]
rootNode = TreeNode(root)

idx = 0
while inorder[idx] != root:
idx += 1
inorder_left = inorder[:idx]
preorder_left = preorder[1:idx + 1]
left_node = self.buildTree(preorder_left, inorder_left)
rootNode.left = left_node

inorder_right = inorder[idx + 1:]
preorder_right = preorder[idx + 1:]
right_node = self.buildTree(preorder_right, inorder_right)
rootNode.right = right_node

return rootNode