[leetcode] 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].

解法一：

类似level order tree traversal，用一个queue存储每一层的node，将每一层的最后一个node，也就是最右的node推到结果中。

/** * 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> res;        if(!root) return res;                queue<TreeNode*> q;        q.push(root);                while(!q.empty()){            int len = q.size();            for(int i=0; i<len; i++){                TreeNode* node = q.front();                q.pop();                if(node->left) q.push(node->left);                if(node->right) q.push(node->right);                if(i==len-1) res.push_back(node->val);            }        }        return res;            }};