[LeetCode] 589. N-ary Tree Postorder Traversal N叉树的后序遍历

 

Given an n-ary tree, return the postorder traversal of its nodes' values.

For example, given a 3-ary tree:

 

 

Return its postorder traversal as: [5,6,3,2,4,1].

 

Note:

Recursive solution is trivial, could you do it iteratively?

 

这道题让我们求N叉树的后序遍历,由于有了之前那道 Binary Tree Postorder Traversal 的基础,了解了二叉树的后序遍历,则N叉树的后序遍历也就没有那么难了。首先还是用递归来做,在递归函数中,判空后,遍历子结点数组,对所有的子结点调用递归函数,然后在 for 循环之外在将当前结点值加入结果 res 数组,这样才能保证是后序遍历的顺序,参见代码如下:

 

解法一:

class Solution {
public:
    vector<int> postorder(Node* root) {
        vector<int> res;
        helper(root, res);
        return res;
    }
    void helper(Node* node, vector<int>& res) {
        if (!node) return;
        for (Node* child : node->children) {
            helper(child, res);
        }
        res.push_back(node->val);
    }
};

 

我们也可以使用迭代的方法来做,这里有个小 trick,写法跟先序遍历十分的像,不同的就是每次把从 stack 中取的结点的值都加到结果 res 的最前面,还有就是遍历子结点数组的顺序是正常的顺序,而前序遍历是从子结点数组的后面往前面遍历,这点区别一定要注意,参见代码如下:

 

解法二:

class Solution {
public:
    vector<int> postorder(Node* root) {
        if (!root) return {};
        vector<int> res;
        stack<Node*> st{{root}};
        while (!st.empty()) {
            Node *t = st.top(); st.pop();
            res.insert(res.begin(), t->val);
            for (Node* child : t->children) {
                if (child) st.push(child);
            }
        }
        return res;
    }
};

 

类似题目:

Binary Tree Postorder Traversal

N-ary Tree Preorder Traversal

N-ary Tree Level Order Traversal

 

参考资料:

https://leetcode.com/problems/n-ary-tree-preorder-traversal/

https://leetcode.com/problems/n-ary-tree-postorder-traversal/discuss/147959/Java-Iterative-and-Recursive-Solutions

 

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2018-10-31 10:11  Grandyang  阅读(2483)  评论(2编辑  收藏  举报
Fork me on GitHub