Binary Tree Level Order Traversal II——LeetCode

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]
]

 题目大意为，按照层次逆序输出每一层的节点，很明显用层次遍历即可，之前在Minimum Depth of Binary Tree ——LeetCode这个题目中，也用到了层次遍历，下面先看我写的，也是用upRow，downRow，来记录上一层、下一层的节点数量，upRow--为0时，一层遍历完了。

　　public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<Integer> level = new LinkedList<>();
        List<List<Integer>> res = new LinkedList<>();
        if (root == null)
            return res;
        ArrayDeque<TreeNode> queue = new ArrayDeque<>();
        queue.add(root);
        int upRow = 1, downRow = 0;
        while (!queue.isEmpty()) {
            TreeNode node = queue.getFirst();
            queue.removeFirst();
            if (node.left != null) {
                queue.add(node.left);
                downRow++;
            }
            if (node.right != null) {
                queue.add(node.right);
                downRow++;
            }
            upRow--;
            level.add(node.val);
            if (upRow == 0) {
                res.add(0, level);
                level = new LinkedList<>();
                upRow = downRow;
                downRow=0;
            }
        }
        return res;
    }

后来发现一种更巧妙的解，就是利用queue的size()，根本不用记录这两个值。

Talk is cheap>>

 public List<List<Integer>> levelOrderBottom2(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        List<List<Integer>> wrapList = new LinkedList<>();

        if (root == null) return wrapList;
        queue.offer(root);
        while (!queue.isEmpty()) {
            int levelNum = queue.size();
            List<Integer> subList = new LinkedList<>();
　　　　　　　//把一层元素全部取出　
            for (int i = 0; i < levelNum; i++) {
                TreeNode node = queue.poll();//取出队列第一个元素
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
                subList.add(node.val);
            }
            wrapList.add(0, subList);
        }
        return wrapList;
    }