[leetcode] 105. Construct Binary Tree from Preorder and Inorder Traversal @ python

最新推荐文章于 2024-01-17 14:50:43 发布

原创最新推荐文章于 2024-01-17 14:50:43 发布 · 156 阅读

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本文介绍了一种通过前序和中序遍历构建二叉树的算法，采用递归方式，利用前序第一个元素为根节点，中序左右划分子树的特性，实现O(n)时间复杂度的解决方案。

原题

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:

/
9 20
/
15 7

解法

递归. 利用preorder的第一个元素是它的根节点, inorder的根节点的左边是它的左子树, 右边是它的右子树这一性质. Base case是当inorder为空时, 直接返回, 然后使用分治法递归.
Time: O(n)
Space: O(1)

代码

# 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
        """
        # base case
        if not inorder:
            return
        
        i = inorder.index(preorder.pop(0))
        root = TreeNode(inorder[i])
        root.left = self.buildTree(preorder, inorder[:i])
        root.right = self.buildTree(preorder, inorder[i+1:])
        return root