题目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].

 分析:就是找到每层最右边的节点

1,深搜

思路: 深搜,一个集合中存放访问到的每层节点,在搜索的过程中,更新对应层的最右边节点

 public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        dfs(root,result,0);
        return result;
     }
     
     private void dfs(TreeNode root, List<Integer> result, int level){
         if(root == null){
             return;
         }
         if(result.size() == level){
             result.add(level,root.val);
         }else{
             result.set(level,root.val); 
         }
         
         dfs(root.left,result,level+1);
         dfs(root.right,result,level+1);
     }

2,利用层次遍历

思路:利用层次遍历, 然后每层遍历结束后,取最右边的节点

 public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        if(root == null){
            return result;
        }
        
        Stack<TreeNode> curLevelNodes = new Stack<TreeNode>();
        curLevelNodes.push(root);
        while(!curLevelNodes.empty()){
            TreeNode rightNode = curLevelNodes.peek();
            result.add(rightNode.val);
            Stack<TreeNode> nextLevelNodes = new Stack<TreeNode>();
            for(TreeNode node : curLevelNodes){
                if(node.left != null){
                    nextLevelNodes.add(node.left);
                }
                if(node.right != null){
                    nextLevelNodes.add(node.right);
                }
            }
          curLevelNodes = nextLevelNodes;
        }
        return result;
    }