Problem

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,
   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---
You should return [1, 3, 4].

思路一(level order traversal)

1. 每一层挑最后一个元素输出

    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode curt = queue.poll();
                if (curt.right != null) {
                    queue.offer(curt.right);
                }
                if (curt.left != null) {
                    queue.offer(curt.left);
                }
                if (i == 0) {
                    result.add(curt.val);
                }
            }
        }
        return result;
    }

思路二(divide & conquer)

1. 假设左右子树分别得到它们的rightsideview,
   1. 如果左子树的list长度比右子树的长, 那么证明我们能从right view 看到一部分左子树

    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        result.add(root.val);
        List<Integer> left = rightSideView(root.left);
        List<Integer> right = rightSideView(root.right);

        int len = Math.max(left.size(), right.size());
        for (int i = 0; i < len; i++) {
            if (i >= right.size()) {
                result.add(left.get(i));
            } else {
                result.add(right.get(i));
            }
        }
        return result;
    }