标签：
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：
标签：
原文地址：http://www.cnblogs.com/sunshisonghit/p/4458594.html
