LeetCode 0590 N-ary Tree Postorder Traversal【N叉树，DFS】-优快云博客

本文探讨了N叉树的后序遍历算法，包括递归和迭代两种实现方式。通过实例展示了输入为特定N叉树时，如何获得节点值的后序遍历结果。约束条件包括树的高度不超过1000，节点总数在0到10^4之间。

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

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Follow up:

Recursive solution is trivial, could you do it iteratively?

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [5,6,3,2,4,1]

Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]

Constraints:

The height of the n-ary tree is less than or equal to 1000
The total number of nodes is between [0, 10^4]

题意

N 叉树后序遍历

思路1

代码1

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<int> postorder(Node* root) {
        vector<int> ans;
        if(root != NULL)
            dfs(root, ans);
        return ans;
    }
    
    void dfs(Node *node, vector<int> &ans){
        if(node == NULL) return;
        for(const auto &item : node->children){
            dfs(item, ans);
        }
        ans.push_back(node->val);
    }
};