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).

Note1

BFS方法：

直接BFS。三层循环的标准形式。

Code1

public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List result = new ArrayList();

        if (root == null) {
            return result;
        }

        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            ArrayList<Integer> level = new ArrayList<Integer>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode head = queue.poll();
                level.add(head.val);
                if (head.left != null) {
                    queue.offer(head.left);
                }
                if (head.right != null) {
                    queue.offer(head.right);
                }
            }
            result.add(level);
        }

        return result;
    }
}

Note2

DFS方法：

其实是pre-order遍历，pre-order开始会一层层向下走，level==res.size()的时候初始化一下内层的列表

Code2

public class Solution {
    /**
     * @param root: A Tree
     * @return: Level order a list of lists of integer
     */
    public List<List<Integer>> levelOrder(TreeNode root) {
        // write your code here
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }

        dfs(root, res, 0);

        return res;
    }

    private void dfs(TreeNode root, List<List<Integer>> res, int level) {
        if (root == null) {
            return;
        }
        if (level == res.size()) {
            res.add(new ArrayList<>());
        }
        res.get(level).add(root.val);

        dfs(root.left, res, level + 1);
        dfs(root.right, res, level + 1);
    }
}