*LeetCode-Binary Tree Right Side View
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应该看出来和level order traversal 一样
第一种方法就是基于这个的 每次展开一层 从最右边开始展开 然后将这一层的第一个放入 即最右边一个 用一个queue维护
public class Solution { public List<Integer> rightSideView(TreeNode root) { Queue<TreeNode> que = new LinkedList<TreeNode> (); List<Integer> res = new ArrayList<Integer>(); if ( root == null ) return res; que.offer( root ); while ( que.size () != 0 ){ int size = que.size(); for ( int i = 0; i < size; i ++ ){ TreeNode cur = que.poll(); if ( i == 0 ) res.add ( cur.val ); if ( cur.right != null ) que.offer( cur.right ); if ( cur.left != null ) que.offer( cur.left ); } } return res; }}
第二种方法就是dfs 很巧妙的就是每次的层数和res里面已经有的个数应该相等
public class Solution { public List<Integer> rightSideView(TreeNode root) { List<Integer> res = new ArrayList<Integer>(); helper ( res, root, 0 ); return res; } public void helper ( List<Integer> res, TreeNode root, int depth ){ if ( root == null ) return; if ( res.size() == depth ) res.add(root.val); helper ( res, root.right, depth + 1); helper ( res, root.left, depth + 1); }}
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