102.BinaryTree LevelOrder Traversal

 

Solution 1:
public List<List<Integer>> levelOrder(TreeNode root) {
Queue<TreeNode> queue = new LinkedList<>();
List<List<Integer>> wrapList = new LinkedList<>();
if(root == null)
return wrapList;
queue.offer(root);
while (!queue.isEmpty()) {
int levelNum = queue.size();
List<Integer> subList = new LinkedList<>();
for(int i = 0; i < levelNum; i++) {
TreeNode n = q.poll();
subList.add(n.val);
if(n.left != null)
q.add(n.left);
if(n.right != null)
q.add(n.right);
}
wrapList.add(subList);
}
return wrapList;
}
    
 
//如果要求把null节点也列出来:
public static List<List<Integer>> levelOrder(TreeNode root) {
    Queue<TreeNode> queue = new LinkedList<>();
    List<List<Integer>> wrapList = new LinkedList<>();
    if(root == null)
        return wrapList;
    queue.offer(root);
    while (!queue.isEmpty()) {
        int levelNum = queue.size();
        List<Integer> subList = new LinkedList<>();
        for(int i = 0; i < levelNum; i++) {
            TreeNode tn = queue.poll();
            if(tn != null)
                queue.offer(tn.left);
            if(tn != null)
                queue.offer(tn.right);
            if (tn == null)
                subList.add(null);
            else
                subList.add(tn.val);
        }
        wrapList.add(subList);
    }
    return wrapList;
}
 
 
Solution 2:
class Solution {
   public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        levelHelper(res, root, 0);
        return res;
    }
    
    public void levelHelper(List<List<Integer>> res, TreeNode root, int height) {
        if (root == null) return;
        if (height >= res.size()) {
            res.add(new LinkedList<>());
        }
        res.get(height).add(root.val);
        levelHelper(res, root.left, height+1);
        levelHelper(res, root.right, height+1);
    }
}

 

posted @ 2019-03-15 14:43  MarkLeeBYR  阅读(89)  评论(0)    收藏  举报