题目:
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]
.
思路分析:
二叉树的层次遍历,在遍历过程中输出每一层最右边的节点。
我们可以在层次遍历过程中先访问右边节点,然后访问做变节点,访问结果入队列,然后每一层最前边的节点放入结果集合中。
C++参考代码:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> rightSideView(TreeNode *root) { vector<int> result; if (!root) return result; queue<TreeNode*> nodeQueue; TreeNode *node = root; nodeQueue.push(node); while (!nodeQueue.empty()) { queue<TreeNode*>::size_type size = nodeQueue.size(); //每次依次取出这一层的节点,然后将改元素的左右节点入队列(如果是第一个节点,则放入结果集合中) for (size_t i = 0; i < size; ++i) { node = nodeQueue.front(); nodeQueue.pop(); if (i == 0) result.push_back(node->val); //注意是先右节点后左节点 if (node->right) nodeQueue.push(node->right); if (node->left) nodeQueue.push(node->left); } } return result; } };
C#参考代码:
/** * Definition for binary tree * public class TreeNode { * public int val; * public TreeNode left; * public TreeNode right; * public TreeNode(int x) { val = x; } * } */ public class Solution { public IList<int> RightSideView(TreeNode root) { IList<int> result = new List<int>(); if (root == null) return result; Queue<TreeNode> nodeQueue = new Queue<TreeNode>(); TreeNode node = root; nodeQueue.Enqueue(node); while (nodeQueue.Count != 0) { int size = nodeQueue.Count; for (int i = 0; i < size; ++i) { node = nodeQueue.Dequeue(); if (i == 0) result.Add(node.val); if (node.right != null) nodeQueue.Enqueue(node.right); if (node.left != null) nodeQueue.Enqueue(node.left); } } return result; } }
Leetcode: Binary Tree Right Side View
原文地址:http://blog.csdn.net/theonegis/article/details/44902247