105. Construct Binary Tree from Preorder and Inorder Traversal

Problem Link

Description

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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 <= 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

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Python3

# 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]:
        N = len(preorder)
        mp = {}
 
        for i, x in enumerate(inorder):
            mp[x] = i
 
        # preorder -> node, left, right
        # inorder -> left, node, right
 
        def go(preIndex, inLeft, inRight):
            if preIndex >= N or inLeft > inRight: return None
 
            node = TreeNode(preorder[preIndex])
            rootIndex = mp[node.val]
 
            node.left = go(preIndex + 1, inLeft, rootIndex - 1)
            node.right = go(preIndex + rootIndex - inLeft + 1, rootIndex + 1, inRight)
 
            return node
        
        return go(0, 0, N - 1)