105.Construct Binary Tree from Preorder and Inorder Traversal

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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:
    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 not len(preorder) or not len(inorder):
            return None
        root = TreeNode(preorder[0])
        pos = inorder.index(root.val)
        inorderleft,inorderright = inorder[:pos],inorder[pos+1:]
        preorderleft,preorderright = preorder[1:1+len(inorderleft)],preorder[1+len(inorderleft):]
        root.left = self.buildTree(preorderleft,inorderleft)
        root.right = self.buildTree(preorderright,inorderright)
        return root

105.Construct Binary Tree from Preorder and Inorder Traversal

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原文地址：https://www.cnblogs.com/bernieloveslife/p/9760758.html