递归解法简单易懂

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
# 求二叉树的右视图
# @param xianxu int整型一维数组 先序遍历
# @param zhongxu int整型一维数组 中序遍历
# @return int整型一维数组
#
class Solution:
    def build_tree(self, preorder, inorder):
        if len(preorder) < 1:
            return None
        root = TreeNode(preorder[0])
        index = inorder.index(preorder[0])
        root.left = self.build_tree(preorder[1:index+1], inorder[:index])
        root.right = self.build_tree(preorder[index+1:], inorder[index+1:])
        return root

    def dfs(self, root, depth):
        if root is None:
            return 

        if depth >= len(self.ans):
            self.ans.append(root.val)
        depth += 1
        self.dfs(root.right, depth)
        self.dfs(root.left, depth)

    def solve(self , xianxu , zhongxu ):
        # write code here
        self.ans = []
        root = self.build_tree(xianxu, zhongxu)
        self.dfs(root, 0)
        return self.ans