leetcode講解--589. N-ary Tree Preorder Traversal

題目

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

For example, given a 3-ary tree:

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

Note:

Recursive solution is trivial, could you do it iteratively?

題目地址

講解

若是用遞歸來解題的話，仍是很是簡單的。若是要用迭代來解題，無非就是用棧來實現。要注意的一點是，須要一個額外的棧來把壓棧的順序倒一下序。

Java代碼

遞歸代碼：

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

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    private List<Integer> result = new ArrayList<>();
    public List<Integer> preorder(Node root) {
        if(root==null){
            return result;
        }
        result.add(root.val);
        for(Node node:root.children){
            preorder(node);
        }
        return result;
    }
}

迭代代碼：

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

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    private List<Integer> result = new ArrayList<>();
    private Stack<Node> stack = new Stack<>();
    private Stack<Node> stack_temp = new Stack();
    public List<Integer> preorder(Node root) {
        if(root==null){
            return result;
        }
        stack.push(root);
        while(!stack.empty()){
            Node theNode = stack.pop();
            result.add(theNode.val);
            for(Node node:theNode.children){
                stack_temp.push(node);
            }
            while(!stack_temp.empty()){
                stack.push(stack_temp.pop());
            }
        }
        return result;
    }
}