105-construct-binary-tree-from-preorder-and-inorder-traversal

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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(Python)

# 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_index = 0
        def array_to_tree(l, r):
            nonlocal preorder_index
            if l > r:
                return None
            n_value = preorder[preorder_index]
            node = TreeNode( preorder[preorder_index])
            preorder_index += 1
            node.left = array_to_tree(l,inorde_index[n_value]-1 )
            node.right = array_to_tree(inorde_index[n_value]+1, r )
            return node
        
        inorde_index = { val:i for i,val in enumerate(inorder)}
        return array_to_tree(0, len(preorder)-1)