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105. Construct Binary Tree from Preorder and Inorder Traversal

Difficulty Topics

Description

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.

Solution

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]:

        def go(preStart, inStart, inEnd):
            if inStart > inEnd or preStart >= len(preorder): return None

            root = TreeNode(preorder[preStart])

            rootIndex = 0
            for index in range(inStart, inEnd + 1):
                if inorder[index] == root.val:
                    rootIndex = index

            root.left = go(preStart + 1, inStart, rootIndex - 1)
            root.right = go(preStart + rootIndex - inStart + 1, rootIndex + 1, inEnd)

            return root

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