105 Construct Binary Tree from Preorder and Inorder Traversal

Original Description
Given preorder and inorder traversal of a tree, construct the binary tree.

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• 要求： 给inorder, preorder traversal 的数组，返回对应的树

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• 返回：
  对应的树

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• 例子：
  • input: preorder = [3,9,20,15,7]， inorder = [9,3,15,20,7]
  • output:
    。 3 。 /
    。 9 20 。。 /
    。。15 7

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Solution

还是recursive的思想，由于从preorder中可以提取root, 可以使用root在inorder的序列分出左右子树，再对左右子树重复上述过程。

# 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 len(preorder) == 0:
            return None
        if len(preorder) == 1:
            return TreeNode(preorder[0])
        
        # recursively use root in preorder to separate inorder into left and right subtree
        root = TreeNode(preorder[0])
        root_idx = inorder.index(root.val)
        root.left = self.buildTree(preorder[1 : root_idx + 1], inorder[0 : root_idx])
        root.right = self.buildTree(preorder[root_idx + 1 :], inorder[root_idx + 1 :])
        return root