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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 <= 3000inorder.length == preorder.length-3000 <= preorder[i], inorder[i] <= 3000preorderandinorderconsist of unique values.- Each value of
inorderalso appears inpreorder. preorderis guaranteed to be the preorder traversal of the tree.inorderis guaranteed to be the inorder traversal of the tree.
# time complexity: O(n) # space complexity: O(n) from typing import List, Optional class Solution: def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]: def dfs(left: int, right: int): nonlocal preorderIndex if left > right: return None rootValue = preorder[preorderIndex] root = TreeNode(rootValue) preorderIndex += 1 root.left = dfs(left, inorderMap[rootValue] - 1) root.right = dfs(inorderMap[rootValue] + 1, right) return root preorderIndex = 0 inorderMap = {} for i, value in enumerate(inorder): inorderMap[value] = i return dfs(0, len(inorder) - 1) preorder = [3, 9, 20, 15, 7] inorder = [9, 3, 15, 20, 7] print(Solution().buildTree(preorder, inorder))
![[Leetcode] 0305. Number of Islands II](https://hogantechs.com/wp-content/uploads/2024/12/13-1024x577.png)
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