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))

原文地址：https://www.cnblogs.com/oldby/p/11184943.html