Construct binary tree from inorder and postorder traversal
# 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 or not inorder:
return None
root_val = postorder[-1]
root = TreeNode(root_val)
idx = inorder.index(root_val)
left_in = inorder[:idx]
right_in = inorder[idx + 1:]
left_post = postorder[:len(left_in)]
right_post = postorder[len(left_in): len(postorder) - 1]
root.left = self.buildTree(left_in, left_post)
root.right = self.buildTree(right_in, right_post)
return root
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:
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 <= 3000postorder.length == inorder.length-3000 <= inorder[i], postorder[i] <= 3000inorderandpostorderconsist of unique values.- Each value of
postorderalso appears ininorder. inorderis guaranteed to be the inorder traversal of the tree.postorderis guaranteed to be the postorder traversal of the tree.