Difficulty

Medium.

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

AC

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


# DFS
class Solution():
    def buildTree(self, inorder, postorder):
        if not len(postorder):
            return
 
 None
        root = TreeNode(postorder[-1])
        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