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