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Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.

 

Example 1:

Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
Output: [3,9,20,null,null,15,7]

Example 2:

Input: inorder = [-1], postorder = [-1]
Output: [-1]

 

Constraints:

  • 1 <= inorder.length <= 3000
  • postorder.length == inorder.length
  • -3000 <= inorder[i], postorder[i] <= 3000
  • inorder and postorder consist of unique values.
  • Each value of postorder also appears in inorder.
  • inorder is guaranteed to be the inorder traversal of the tree.
  • postorder is guaranteed to be the postorder traversal of the tree.

 

My 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 buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        if not inorder or not postorder:
            return None
        
        root_val = postorder[-1]
        root = TreeNode(root_val)
        if len(inorder) == 1:
            return root
        
        root_idx = inorder.index(root_val)

        root.left = self.buildTree(inorder[:root_idx], postorder[:root_idx])
        root.right = self.buildTree(inorder[root_idx + 1:], postorder[root_idx:-1])

        return root

 

 

posted on 2025-03-26 09:09  ZhangZhihuiAAA  阅读(10)  评论(0)    收藏  举报