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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].
这道题就是BT的Level Order Traversal,每次要换一层的时候,记录当前节点
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<Integer> rightSideView(TreeNode root) { 12 ArrayList<Integer> res = new ArrayList<Integer>(); 13 if (root == null) return res; 14 LinkedList<TreeNode> queue = new LinkedList<TreeNode>(); 15 queue.offer(root); 16 int PNum = 1; 17 int CNum = 0; 18 while (!queue.isEmpty()) { 19 TreeNode cur = queue.poll(); 20 PNum--; 21 if (cur.left != null) { 22 queue.offer(cur.left); 23 CNum++; 24 } 25 if (cur.right != null) { 26 queue.offer(cur.right); 27 CNum++; 28 } 29 if (PNum == 0) { 30 res.add(cur.val); 31 PNum = CNum; 32 CNum = 0; 33 } 34 } 35 return res; 36 } 37 }
Leetcode: Binary Tree Right Side View
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原文地址:http://www.cnblogs.com/EdwardLiu/p/4416160.html