[LeetCode] 106. Construct Binary Tree from Postorder and Inorder Traversal_Medium tag: Tree Traversal

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

这个题目思路跟[LeetCode] 105. Construct Binary Tree from Preorder and Inorder Traversal_Medium tag: Tree Traversal一样, 只是root是postorder[-1], 而inorder[0] 而已, 本质一样.

 

Code:

class Solution:
    def buildTree(self, postorder, inorder):
        if not postorder or not inorder: return
        root, index = TreeNode(postorder[-1]), inorder.index(postorder[-1])
        root.left = self.buildTree(postorder[:index], inorder[:index])
        root.right = self.buildTree(postorder[index: -1], inorder[index+1:])
        return root