106. 从中序与后序遍历序列构造二叉树

https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

解法一：递归

和105题类似，区别在于后序的-1号元素是中序的根，划分之后递归

class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
        if len(inorder) == 0 or len(postorder) == 0:
            return
        root = TreeNode(postorder[-1])
        mid = inorder.index(postorder[-1])
        #以mid为界划分中序
        in_l = inorder[:mid]
        in_r = inorder[mid+1:]
        #以mid为界划分后序
        post_l = postorder[:len(in_l)]
        post_r = postorder[len(in_l):-1]
        #递归构造左右子树
        root.left = self.buildTree(in_l, post_l)
        root.right = self.buildTree(in_r, post_r)
        return root