Construct Binary Tree from Preorder and Inorder Traversal

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:
# 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 getRoot(self, p, q):
        return min((i for i in range(p, q)), key = lambda x: self.valIndex[self.inorder[x]])
    
    def buildTreeRec(self, p, q):
        if p < q:
            root = TreeNode()
            currRoot = self.getRoot(p, q)
            root.val = self.inorder[currRoot]
            if currRoot > p:
                root.left = self.buildTreeRec(p, currRoot)
            if currRoot < q - 1:
                root.right = self.buildTreeRec(currRoot + 1, q)
            return root
    
    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        n = len(inorder)
        self.inorder = inorder
        self.valIndex = {}
        for i, item in enumerate(preorder):
            self.valIndex[item] = i
        return self.buildTreeRec(0, n)

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