107. Binary Tree Level Order Traversal II

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Total Accepted: 70095
Total Submissions: 211952
Difficulty: Easy

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.

Java:

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/**
 * 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 LinkedList<List<Integer>>();            
        if(root==null)
            return res;
        Queue<TreeNode> q =new LinkedList<TreeNode>();
        q.offer(root);
        while(!q.isEmpty()){
            int size=q.size();
            List<Integer> tmp=new LinkedList<Integer>();                
            for(int i=0;i<size;++i){
                TreeNode temp=q.poll();
                if(temp.left!=null)
                    q.offer(temp.left);
                if(temp.right!=null)
                    q.offer(temp.right);
                tmp.add(temp.val);
            }
            res.add(0, tmp);
        }
        return res;
    }
}

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