106 Construct Binary Tree from Inorder and Postorder Traversal

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:

  3
 / \
9   20
   /  \
  15   7

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.

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# 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 postorder:
            return None
        root_val = postorder[-1]
        root = TreeNode(root_val)
        idx = inorder.index(root_val)
        root.left = self.buildTree(inorder[:idx], postorder[:idx])
        root.right = self.buildTree(inorder[idx+1:], postorder[idx:-1])
        return root