107. Binary Tree Level Order Traversal II Java Solutions

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).

For 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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        Stack<List<Integer>> stack = new Stack<List<Integer>>();
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        if(root == null) return res;
        q.add(root);
        while(!q.isEmpty()){
            List<Integer> tmp = new ArrayList<Integer>();
            int size = q.size();
            for(int i = 0; i< size; i++){
                TreeNode node = q.poll();
                tmp.add(node.val);
                if(node.left != null) q.add(node.left);
                if(node.right != null) q.add(node.right);
            }
            stack.push(tmp);
            
        }
        while(!stack.isEmpty()){
            res.add(stack.pop());
        }
        return res;
    }
}