Problem
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
Solution
The problem is almost the same as 105 Construct BT from Preorder and Inorder. Notes can be found there.
class Solution:
def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
if inorder:
node = TreeNode(postorder.pop())
split = inorder.index(node.val)
node.right = self.buildTree(inorder[split + 1:], postorder)
node.left = self.buildTree(inorder[:split], postorder)
return node