105. 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]

從前中序結果建出原本的樹，前序可以找root, 中序可以建樹

T: O(n), S: O(1)
# 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]) -> TreeNode:
        # pre : (root, left, right) root is first, find root
        # in: (left, root, right) construct tree

        # Check if the root exist in inorder
             if len(preorder) and preorder[0] in inorder:
            
            # Find the root index in inorer
            root_index = inorder.index(preorder[0])
        
            # Make the root node
            tree = TreeNode(preorder[0])

            # Build left subtree
            tree.left = self.buildTree(preorder[1:root_index+1], inorder[:root_index])

            # Build right subtree
            tree.right = self.buildTree(preorder[root_index+1:], inorder[root_index+1:])
            return tree

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