BinaryTreeLevelOrderTraversal.java

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

/**
 * Binary Tree Level Order Traversal http://oj.leetcode.com/problems/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,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3],
 * [9,20], [15,7] ]
 * 
 * @author joshluo
 * 
 */

public class BinaryTreeLevelOrderTraversal {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if (root == null)
            return result;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        ArrayList<Integer> currentLevel = new ArrayList<Integer>();
        int currentLevelNodes = 1, nextLevelNodes = 0;
        while (queue.size() > 0) {
            TreeNode current = queue.remove();
            currentLevelNodes--;
            currentLevel.add(current.val);
            if (current.left != null) {
                queue.add(current.left);
                nextLevelNodes++;
            }
            if (current.right != null) {
                queue.add(current.right);
                nextLevelNodes++;
            }
            if (currentLevelNodes == 0) {
                currentLevelNodes = nextLevelNodes;
                nextLevelNodes = 0;
                result.add(currentLevel);
                currentLevel = new ArrayList<Integer>();
            }
        }
        return result;
    }
}