LintCode: Binary Tree Level Order Traversal

Problem:

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

Sample Input:

Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

Sample Output:

return its level order traversal as:

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

Idea:

The idea is to set a queue, the size for loop is very important.

Code:

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
 
 
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Level order a list of lists of integer
     */
    // set the result variable
    private ArrayList result = new ArrayList();
    // set the queue
    private Queue<TreeNode> queue = new LinkedList<TreeNode>();
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        // write your code here
        // handle corner cases
        if (root == null) {
            return result;
        }
        // add root to the queue first
        queue.offer(root);
        // when queue is not empty, execute
        while (!queue.isEmpty()) {
            ArrayList<Integer> temp = new ArrayList<Integer>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                temp.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            result.add(temp);
        }
        return result;
    }
}