[LeetCode] 589. N-ary Tree Preorder Traversal (vs.

array_huang 发布于2019-08-19 11:18 / 2999人阅读

589. N-ary Tree Preorder Traversal

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?

Solution (Iteration)

Using stack, push the child from the end of list

class Solution {
    public List preorder(Node root) {
        List res = new ArrayList<>();
        if (root == null) return res;
        Stack stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            Node cur = stack.pop();
            res.add(cur.val);
            for (int i = cur.children.size()-1; i >= 0; i--) stack.push(cur.children.get(i));
        }
        return res;
    }
}

Solution (Recursion)

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

    public Node() {}

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

144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes" values.

Example:

Input: [1,null,2,3]

Output: [1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?

Solution (Iteration)

Use stack, first push node.right, then push node.left

class Solution {
    public List preorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        if (root == null) return res;
        Stack stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            res.add(cur.val);
            if (cur.right != null) stack.push(cur.right);
            if (cur.left != null) stack.push(cur.left);
        }
        return res;
    }
}

Solution (Recursion)

class Solution {
    public List preorderTraversal(TreeNode root) {
        List res = new ArrayList<>();
        helper(root, res);
        return res;
    }
    private void helper(TreeNode root, List res) {
        if (root == null) return;
        res.add(root.val);
        helper(root.left, res);
        helper(root.right, res);
    }
}

云服务器服务器托管 Preorder Traversal LeetCode LeetCode_String

文章版权归作者所有，未经允许请勿转载,若此文章存在违规行为，您可以联系管理员删除。

转载请注明本文地址：https://www.ucloud.cn/yun/77377.html

Leetcode PHP题解--D43 589. N-ary Tree Preorder Trave

摘要：题目链接题目分析维数组的先序遍历。这题也不想多说什么了。是比较基础的题目了。先序就是先根后子而已。思路在遍历子节点之前，先保存当前节点的信息。最终代码若觉得本文章对你有用，欢迎用爱发电资助。 D43 589. N-ary Tree Preorder Traversal 题目链接 589. N-ary Tree Preorder Traversal 题目分析 N维数组的先序遍历。这题也...

junbaor 2019-07-01 12:43 评论0 收藏0
[LeetCode] 590. N-ary Tree Postorder Traversal (vs

摘要：按顺序放入，正好方面是从到，顺序方面是从最右到最左，因为是先入后出。这样最后一下就是先左后右，先子后根。 590. N-ary Tree Postorder Traversal Problem Given an n-ary tree, return the postorder traversal of its nodes values.For example, given a 3-ar...

sydMobile 2019-08-19 11:18 评论0 收藏0
[LeetCode] 429. N-ary Tree Level Order Traversal (

429. N-ary Tree Level Order Traversal Given an n-ary tree, return the level order traversal of its nodes values. (ie, from left to right, level by level). For example, given a 3-ary tree:showImg(https...

LiangJ 2019-08-19 11:18 评论0 收藏0
leetcode429. N-ary Tree Level Order Traversal

摘要：题目要求对叉树进行水平遍历，并输出每一行遍历的结果。因此无需再用队列来额外存储每一行的水平遍历，可以直接通过递归将遍历结果插入到相应行的结果集中。题目要求 Given an n-ary tree, return the level order traversal of its nodes values. (ie, from left to right, level by level)...

tomlingtm 2019-08-16 16:57 评论0 收藏0
Leetcode PHP题解--D55 429. N-ary Tree Level Order Tr

摘要：题目链接题目分析按层遍历叉树。思路以层数为键，塞入当前节点的值。最终代码若觉得本文章对你有用，欢迎用爱发电资助。 D55 429. N-ary Tree Level Order Traversal 题目链接 429. N-ary Tree Level Order Traversal 题目分析按层遍历N叉树。思路以层数为键，塞入当前节点的值。递归遍历即可。最终代码

libxd 2019-07-01 12:45 评论0 收藏0