33 N-ary Tree Postorder Traversal

題目

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:

Example 2:

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]

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

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

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

分析

題意：n叉樹的後序遍歷

思路：
只需要利用遍歷“最小元”的思想進行遞歸即可。

n叉樹後序遍歷的最小元：遍歷其他節點–>遍歷根節點
從題意可知，根節點是root，其他節點是root.children中的節點。
因此算法就是，先遞歸遍歷root.children的所有節點，再遍歷根節點

解答

class Solution {
	// 存放結果
    List<Integer> list = new ArrayList<>();
    public List<Integer> postorder(Node root) {
    	// 遞歸終點
        if (root == null)
            return list;
        // 遞歸遍歷其他節點
        for(Node node: root.children)
            postorder(node);
        // 遍歷根節點
        list.add(root.val);
        return list;
    }
}