leetcode-106-从中序和后序遍历构造二叉树

题目描述：

 

方法一：O(n) O(n)

class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode:
        assert len(inorder)==len(postorder)
        if len(inorder)==0:
            return None
        if len(inorder)==1:
            return TreeNode(inorder[0])
        pos = inorder.index(postorder[-1])
        root = TreeNode(postorder[-1])
        root.left = self.buildTree(inorder[:pos],postorder[:pos])
        root.right = self.buildTree(inorder[pos+1:],postorder[pos:-1])
        return root

 方法二;

class Solution: 
    def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode: 
        d={j:i for i,j in enumerate(inorder)} 
        def f(i,j): 
            if i<j: 
                t=TreeNode(postorder.pop()) 
                t.right=f(d[t.val]+1,j) 
                t.left=f(i,d[t.val]) 
                return t 
        return f(0,len(inorder))