题意

实现二叉树的右视图

方法

BFS

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> ans;
        if(!root) return ans;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty())
        {
            int n = q.size();
            for(int i = 0; i < n; i++)
            {
                auto node = q.front();
                q.pop();
                if(i == n-1) //记录每层中最后一个节点的值
                {
                    ans.push_back(node->val);
                }
                if(node->left) q.push(node->left);
                if(node->right) q.push(node->right);
            }
        }
        return ans;
    }
};

原文地址：http://www.cnblogs.com/Figure_at_a_Window/p/16913057.html