107. Binary Tree Level Order Traversal II

 

Solution1:广度优先

class Solution {

    public List<List<Integer>> levelOrderBottom(TreeNode root) { 

        Queue<TreeNode> q = new LinkedList<>();

        List<List<Integer>> wrap = new LinkedList<>();

        if (root == null) return wrap;

        q.offer(root);

        while (!q.isEmpty()) {

           int num = q.size();

           List<Integer> sub = new LinkedList<>();

           for (int i = 0; i < num; i++) {

               TreeNode n1 = q.poll();

               if (n1.left != null) q.offer(n1.left); 

               if (n1.right != null) q.offer(n1.right);

               sub.add(n1.val);

           }

           wrap.add(0, sub); 

        }

       return wrap;

    }

}

Solution2:深度优先

class Solution {

    public List<List<Integer>> levelOrderBottom(TreeNode root) {

         List<List<Integer>> wrap = new LinkedList<>();

         LevelMaker(wrap, root, 0);

         return wrap;

    }

    public void LevelMaker(List<List<Integer>> list, TreeNode root, int level) {

         if (root == null) return;

         if (level >= list.size()) {

            list.add(0, new LinkedList<>()); //先创建空的链表，最后再加入值

         }

         LevelMaker(list, root.left, level + 1);

         LevelMaker(list, root.right, level + 1);

         list.get(list.size() - 1 - level).add(root.val); //level和对应链表的位置：2-0；1-1；0-2，和为list.size()-1

    }   

}