LeetCode106——Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2022-05-29 13:27:49 发布

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博客围绕根据树的中序和后序遍历构建二叉树展开，给出题目链接，假设树中无重复元素，并给出示例，还展示了代码及运行结果。

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题目链接

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

代码

class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
        if len(inorder)!=len(postorder) or len(inorder) == 0:
            return None
        if len(inorder) == 1:
            return TreeNode(postorder[-1])

        root = TreeNode(postorder[-1])
        index = inorder.index(root.val)
        root.left = self.buildTree(inorder[0:index], postorder[0:index])
        root.right = self.buildTree(inorder[index+1:], postorder[index:-1])
        return root

运行结果

Runtime: 216 ms, faster than 44.97% of Python3 online submissions for Construct Binary Tree from Inorder and Postorder Traversal.
Memory Usage: 87.6 MB, less than 8.55% of Python3 online submissions for Construct Binary Tree from Inorder and Postorder Traversal.