199. Binary Tree Right Side View

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].

代码如下：

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public List<Integer> rightSideView(TreeNode root) {
         boolean flag=true;
         List<Integer> list=new ArrayList<>();
         ArrayList<TreeNode> res=new ArrayList<TreeNode>();
         Map<Integer,Integer> map=new HashMap<>();

         if(root==null)
         return list;

          res=LevelTraverse(root);
          list.add(root.val);
          map.put(0,0);

         while(flag)
         {
        while(root.right!=null)
        {
            root=root.right;
            list.add(root.val);
            map.put(res.indexOf(root),res.indexOf(root));

        }
          if(root.left!=null)
        {
            root=root.left;
            list.add(root.val);
            map.put(res.indexOf(root),res.indexOf(root));
        }
        else{
             if(res.indexOf(root)-1<0)
             flag=false;
             else {
                 root=res.get(res.indexOf(root)-1);
                 if(map.containsKey(res.indexOf(root)))
                 break;
             }
        }
         }
         return list;
     }

     public ArrayList<TreeNode> LevelTraverse(TreeNode root)//树的水平遍历
     {
         ArrayList<TreeNode> list=new ArrayList<TreeNode>();
         Queue<TreeNode> queue = new LinkedList<TreeNode>();
         queue.add(root);
         while(queue.size()>0)
         {
             TreeNode a=(TreeNode)queue.peek();
             queue.remove();
             list.add(a);
             if(a.left!=null)
             queue.add(a.left);
             if(a.right!=null)
             queue.add(a.right);
         }

         return list;
     }
 }