102_BinaryTreeLevelOrderTraversal102.java

/**
 * 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]
 * ]
 */


/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */


import java.util.LinkedList;
import java.util.Queue;
import java.util.ArrayList;
import java.util.List;


public class BinaryTreeLevelOrderTraversal102 {
    public List<List<Integer>> levelOrder(TreeNode root) {

        Queue<TreeNode> q = new LinkedList<>();
        List<List<Integer>> result = new ArrayList<>();
        int level = 0;

        if (root == null) {
            return result;
        } else {
            q.add(root);
        }

        helper(q, result, 0);

        return result;
    }


    private void helper(Queue<TreeNode> q, List<List<Integer>> result, int level) {
        int s = q.size();
        if (s == 0) {
            return;
        }

        addLevel(result, level);

        int i = 0;
        while (i < s) {
            TreeNode now = q.poll();
            result.get(level).add(now.val);
            if (now.left != null) {
                q.add(now.left);
            }
            if (now.right != null) {
                q.add(now.right);
            }
            i++;
        }

        helper(q, result, level + 1);
    }


    private void addLevel(List<List<Integer>> result, int level) {
        while (result.size() <= level) {
            result.add(new ArrayList<Integer>());
        }
    }


    /**
     * https://discuss.leetcode.com/topic/7647/java-solution-with-a-queue-used
     */
    public List<List<Integer>> levelOrder2(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();

        if(root == null) return wrapList;

        queue.offer(root);
        while(!queue.isEmpty()){
            int levelNum = queue.size();
            List<Integer> subList = new LinkedList<Integer>();
            for(int i=0; i<levelNum; i++) {
                if(queue.peek().left != null) queue.offer(queue.peek().left);
                if(queue.peek().right != null) queue.offer(queue.peek().right);
                subList.add(queue.poll().val);
            }
            wrapList.add(subList);
        }
        return wrapList;
    }


    public List<List<Integer>> levelOrder3(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            List<Integer> level = new ArrayList<>();
            int cnt = queue.size();
            for (int i = 0; i < cnt; i++) {
                TreeNode node = queue.poll();
                level.add(node.val);
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
            res.add(level);
        }
        return res;
    }

    /**
     * https://discuss.leetcode.com/topic/7332/java-solution-using-dfs
     */

    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        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<Integer>());
        }
        res.get(height).add(root.val);
        levelHelper(res, root.left, height+1);
        levelHelper(res, root.right, height+1);
    }

}