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105. Construct Binary Tree from Preorder and Inorder Traversal ​

题目 ​

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

 

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

 

Constraints:

  • 1 <= preorder.length <= 3000
  • inorder.length == preorder.length
  • -3000 <= preorder[i], inorder[i] <= 3000
  • preorder and inorder consist of unique values.
  • Each value of inorder also appears in preorder.
  • preorder is guaranteed to be the preorder traversal of the tree.
  • inorder is guaranteed to be the inorder traversal of the tree.
Related Topics
  • 树
  • 数组
  • 哈希表
  • 分治
  • 二叉树

  • 👍 2557
  • 👎 0

    • 思路 ​

    dfs 通过遍历 tree

    • 通过 hashmap 存储 inorder 值的位置，将查找复杂度由 O(N) -> O（1）
    • 递归时 通过记录 索引，而不是拆分数组，可降低空间消耗
    • 函数内使用到外部的变量时需要手动声明
      • nolocal 声明变量

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution:
    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        in_dict = {
        pre_index = 0

        for i in range(len(inorder)):
            in_dict[inorder[i]] = i


        def buildSubTree(in_left, in_right) -> Optional[TreeNode]:
            nonlocal pre_index

            if in_left > in_right:
                return None

            root = TreeNode(preorder[pre_index])
            pre_index += 1

            root_index = in_dict[root.val]
            root.left = buildSubTree(in_left, root_index - 1)
            root.right = buildSubTree(root_index + 1, in_right)

            return root


        return buildSubTree(0, len(inorder) - 1)




        
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(N)