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;

    }