[leetcode]Binary Tree Level Order Traversal II

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).

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]
] 

算法思路：

典型的BFS，与[leetcode]Binary Tree Level Order Traversal完全一样，不过一个是对list的头插，一个是尾插法。

代码如下：

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>();
        Queue<TreeNode> copy = new LinkedList<TreeNode>();
        q.offer(root);
        res.add(new ArrayList<Integer>(Arrays.asList(root.val)));
        while(!q.isEmpty()){
            TreeNode node = q.poll();
            if(node.left != null) copy.offer(node.left);
            if(node.right != null) copy.offer(node.right);
            if(q.isEmpty() && !copy.isEmpty()){
                List<Integer> list = new ArrayList<Integer>();
                while(!copy.isEmpty()){
                    q.offer(copy.peek());
                    list.add(copy.poll().val);
                }
                res.add(0,list);
            }
        }
        return res;
    }
}