[LeetCode] 429. N-ary Tree Level Order Traversal (

LiangJ 發布于2019-08-19 11:18 / 2612人閱讀

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:

We should return its level order traversal:

[
     [1],
     [3,2,4],
     [5,6]
]

Note:

The depth of the tree is at most 1000.
The total number of nodes is at most 5000.

Solution (Iteration)

use Queue, in each level, use queue.size() to loop

class Solution {
    public List> levelOrder(Node root) {
        List> res = new ArrayList<>();
        if (root == null) return res;
        Queue queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List curRes = new ArrayList<>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                Node node = queue.poll();
                curRes.add(node.val);
                for (Node child: node.children) {
                    queue.offer(child);
                }
            }
            res.add(curRes);
        }
        return res;
    }
}

102. Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes" values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

Solution (Iteration)

class Solution {
    public List> levelOrder(TreeNode root) {
        List> res = new ArrayList<>();
        if (root == null) return res;
        Queue queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List curRes = new ArrayList<>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                curRes.add(node.val);
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            res.add(curRes);
        }
        return res;
    }
}

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