[Leetcode/파이썬] 105. Construct Binary Tree from Preorder and Inorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal

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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.

 

# 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]:
        inorder_map = {val: idx for idx, val in enumerate(inorder)}
        self.idx = 0

        def build(start, end) :
            if start > end :
                return None 

            root_val = preorder[self.idx]
            self.idx += 1
            root = TreeNode(root_val)

            mid = inorder_map[root_val]

            root.left = build(start, mid - 1)
            root.right = build(mid + 1, end)

            return root

        return build(0, len(inorder) - 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]) -> Optional[TreeNode]:
        if preorder and inorder :
            val = preorder.pop(0)
            root = TreeNode(val)
            
            root.left = self.buildTree(preorder, inorder[:inorder.index(val)])
            root.right = self.buildTree(preorder, inorder[inorder.index(val)+1:] )

            return root

        return None

이런 빌딩하는 문제들 조금 더 많이 풀어봐야겠다. 코딩테스트에는 별 도움 안될지는 몰라도 실제로 작업하는 능력에는 도움이 많이 될 거 같다.