LeetCode 199 Binary Tree Right Side View（二叉树层序遍历）

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].

题目大意：给出一棵二叉树，假设你现在站在它的右侧，输出你看到的节点（即二叉树的右视图）。

解题思路：二叉树层序遍历，输出每一层的最后一个节点即可。因为层序遍历有递归和非递归两种写法，所以本题也给出了两种解法。

代码如下：

/** * 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) {        // levelOrderTraversal(root, ans);        _levelOrderTraversal(root, 0, ans);        return ans;    }private:    queue<TreeNode*> que;    vector<int> ans;    //非递归版本    void levelOrderTraversal(TreeNode* root, vector<int>& ans) {        if(root == nullptr) return ;        int cur = 0, last, val;        TreeNode* tmp;        que.push(root);                while(cur < que.size()) {            last = que.size();            while(cur < last) {                tmp = que.front();                que.pop();                if(tmp->left) que.push(tmp->left);                if(tmp->right) que.push(tmp->right);                cur++;            }            ans.push_back(tmp->val);            cur = 0;        }    }    //递归版本    void _levelOrderTraversal(TreeNode* root, int depth, vector<int>& ans) {        if(root == nullptr) return ;        if(depth >= ans.size()) ans.push_back(root->val);        _levelOrderTraversal(root->right, depth + 1, ans);        _levelOrderTraversal(root->left, depth + 1, ans);    }};