Binary Tree Right Side View

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;

    }

}

View Code

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);



    }

}

View Code

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。