Binary Tree Right Side View

问题：

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,

思路：

　　bfs+dfs

我的代码：

public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> rst = new ArrayList<Integer>();
        if(root == null) return rst;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while(!queue.isEmpty())
        {
            int size = queue.size();
            for(int i = 0; i < size; i++)
            {
                TreeNode node = queue.poll();
                if(i == size-1)
                {
                    rst.add(node.val);
                }
                if(node.left != null)
                    queue.offer(node.left);
                if(node.right != null)
                    queue.offer(node.right);
            }
        }
        return rst;
    }
}

他人代码：

public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        rightView(root, result, 0);
        return result;
    }

    public void rightView(TreeNode curr, List<Integer> result, int currDepth){
        if(curr == null){
            return;
        }
        if(currDepth == result.size()){
            result.add(curr.val);
        }

        rightView(curr.right, result, currDepth + 1);
        rightView(curr.left, result, currDepth + 1);

    }
}

学习之处：

• 没有想到DFS的解法啊，看到别人怎么写的，思路真实精妙啊，可以通过层数控制只输出一侧。