资讯专栏INFORMATION COLUMN

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

array_huang / 3157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]

   1
    
     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);
    }
}

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

相关文章

  • 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 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 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 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 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 评论0 收藏0

发表评论

0条评论

最新活动
阅读需要支付1元查看
<