[LeetCode]--107. 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,null,null,15,7],

  3
 / \
 9  20
 /  \
15   7

return its bottom-up level order traversal as:

 [
   [15,7],
   [9,20],
   [3]
 ]

想着用之前顺序的方法，但是List在add的时候用栈来存放，然后再pop出来就逆序了。

public List<List<Integer>> levelOrder1(TreeNode root) {
        List<List<Integer>> list = new ArrayList<List<Integer>>();
        if (root == null)
            return list;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        Stack<List<Integer>> stack = new Stack<List<Integer>>();
        queue.add(root);
        while (!queue.isEmpty()) {
            List<Integer> list1 = new ArrayList<Integer>();
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode t1 = queue.poll();
                list1.add(t1.val);
                if (t1.left != null)
                    queue.add(t1.left);
                if (t1.right != null)
                    queue.add(t1.right);
            }
            //list.add(list1);
            stack.push(list1);
        }
        while (!stack.isEmpty()) {
            list.add(stack.pop());
        }
        return list;
    }

也是一次就通过了，再去看看别人的代码。

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: buttom-up level order a list of lists of integer
     */
    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            ArrayList<Integer> level = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode head = queue.poll();
                level.add(head.val);
                if (head.left != null) {
                    queue.offer(head.left);
                }
                if (head.right != null) {
                    queue.offer(head.right);
                }
            }
            result.add(level);
        }

        Collections.reverse(result);
        return result;
    }
}

最后用系统自带的reverse方法逆转也是可行的。