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105. Construct Binary Tree From Preorder And Inorder Traversal

Array Hash Table Divide and Conquer Tree Binary Tree

Problem - Construct Binary Tree From Preorder And Inorder Traversal

Medium

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.

Solutions

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# 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]:
        def dfs(i: int, j: int, n: int) -> Optional[TreeNode]:
            if n <= 0:
                return None
            val = preorder[i]
            k = hash_table[val]
            left = dfs(i + 1, j, k - j)
            right = dfs(i + 1 + k - j, k + 1, n - k + j - 1)

            return TreeNode(val, left, right)

        hash_table = {val: i for i, val in enumerate(inorder)}
        return dfs(0, 0, len(preorder))

Submission Stats:

  • Runtime: 7 ms (63.29%)
  • Memory: 19.3 MB (71.81%)