[LeetCode] 429. N-ary Tree Level Order Traversal (vs. LC102)_Node.js

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[LeetCode] 429. N-ary Tree Level Order Traversal (vs. LC102)

发布时间：2019-06-16 发布网站：脚本宝典

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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

      
      
      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;
    }
}" title="" data-original-title="复制">
      
      
class Solution {
    public List<List<Integer>> levelOrder(Node root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Queue<Node> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> 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],

      
      
      
      
      
    3
   / 
  9  20
    /  
   15   7
return its level order traversal as:

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

Solution (Iteration)

      
      
      
      
      
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            List<Integer> 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;
    }
}