Binary Tree Level Order Traversal II.java

Compared with BFS Template. Just reverse the result then achieve the goal.

/*
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).

Example
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]
]

Tags Expand 
Queue Binary Tree Binary Tree Traversal Breadth First Search
*/

/**
 * 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 List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> rst = new ArrayList<>();
        
        if (root == null) {
            return rst;
        }
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            ArrayList<Integer> level = new ArrayList<Integer>();
            for (int i = 0; i < size; 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);
                }
            }
            rst.add(level);
        }
        
        Collections.reverse(rst);
        return rst;
    }
}