105. Construct Binary Tree from Preorder and Inorder Traversal

Medium (Trung bình) Python 🔗 Xem trên LeetCode

📋 Đề Bài

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.

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Tree Traversal (Duyệt cây)
⏱️ Thời gian O(V+E)
💾 Không gian O(V)

💻 Lời Giải

Python 0105-construct-binary-tree-from-preorder-and-inorder-traversal.py
# 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]:
        preorder = preorder[::-1]
        
        def dfs(preorder, inorder):
            if not inorder:
                return None
            idx = inorder.index(preorder.pop())
            node = TreeNode(inorder[idx])
            node.left = dfs(preorder, inorder[:idx])
            node.right = dfs(preorder, inorder[idx + 1:])
            return node
        
        return dfs(preorder, inorder)