[LeetCode]199. Binary Tree Right Side View

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].

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思路：二叉树层序遍历，只提取每层的最后一个元素。

/** * Definition for a binary tree node. * 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 == NULL)            return result;        queue<TreeNode*> q;        q.push(root);                   // root不为空，压入队列        int toBePrint = 1;              // 记录本层还需要打印的节点个数        int nextLevel = 0;              // 记录下一层的节点个数，每次入队列加1        vector<int> oneLayer;           // 单层节点        while(!q.empty()){            TreeNode* pCur = q.front();            q.pop();            oneLayer.push_back(pCur->val);            if(pCur->left){             // 左节点不为空，入队列                q.push(pCur->left);                ++nextLevel;            }            if(pCur->right){            // 右结点不为空，入队列                q.push(pCur->right);                ++nextLevel;            }            if(--toBePrint == 0){       // toBePrint 为0，则将oneLayer压入result                toBePrint = nextLevel;                nextLevel = 0;                result.push_back(*(oneLayer.end()-1));                oneLayer.clear();       // oneLayer清空            }        }        return result;    }};

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