[LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal

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

题目

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

思路

递归，建树。
还需要改进

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

def tbuild(preorder,inorder,pstart,pend,istart,iend):
    if pstart >= pend:
        return None
    node = TreeNode(preorder[pstart])

    for i in range(istart,iend):
        if inorder[i] == node.val:
            break
    print(preorder[pstart])
    node.left = tbuild(preorder,inorder,pstart + 1,pstart+(i - istart)+1,istart,i)
    node.right = tbuild(preorder,inorder,pstart+(i - istart)+1,pend,i+1,iend)
    return node


class Solution:
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        return tbuild(preorder, inorder,0,len(preorder),0,len(inorder))