[刷题]Binary Tree Level Order Traversal II

/**
* 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: buttom-up level order a list of lists of integer
*/
public ArrayList<ArrayList<Integer>> levelOrderButtom(TreeNode root) {
// 2015-3-26 BFS
ArrayList<ArrayList<Integer>> rst = new ArrayList<>();
if (root == null) {
return rst;
}

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

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

return rst;
}
}