lintcode-medium-Binary Tree Level Order Traversal II

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

Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its bottom-up level order traversal as:

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

/**
 * 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>> levelOrderBottom(TreeNode root) {
        // write your code here
        
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        
        if(root == null)
            return result;
        
        ArrayList<Integer> line = new ArrayList<Integer>();
        Queue<TreeNode> curr = new LinkedList<TreeNode>();
        Queue<TreeNode> next = new LinkedList<TreeNode>();
        
        curr.offer(root);
        
        while(!curr.isEmpty()){
            TreeNode temp = curr.poll();
            
            line.add(temp.val);
            
            if(temp.left != null)
                next.offer(temp.left);
            if(temp.right != null)
                next.offer(temp.right);
            
            if(curr.isEmpty()){
                result.add(0, new ArrayList<Integer>(line));
                curr = next;
                next = new LinkedList<TreeNode>();
                line = new ArrayList<Integer>();
            }
            
        }
        return result;
    }
}