力扣199. 二叉树的右视图

广度优先搜索

• 思路：
  • 使用广度优先搜索，那么每层最后一次遍历的节点就是最右视图的节点；
  • 广度优先搜索模板：

    • std::queue nq;
      nq.push(root);
      
      while (!nq.empty()) {
          int levelSize = nq.size();
                  
          // deal with every layer
          for (int idx = 0; idx < levelSize; ++idx) {
              auto node = nq.front();
              nq.pop();
      
              
              /// do something
              // if (idx == levelSize - 1) {
              //     result.push_back(node->val);
              // }
      
              if (node->left) {
                  nq.push(node->left);
              }
              if (node->right) {
                  nq.push(node->right);
              }
          }
      }

• 完整代码：

/**
 * 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 rightSideView(TreeNode* root) {
        std::vector result;
        if (root == nullptr) {
            return result;
        }

        std::queue nq;
        nq.push(root);

        while (!nq.empty()) {
            int levelSize = nq.size();
            
            for (int idx = 0; idx < levelSize; ++idx) {
                auto node = nq.front();
                nq.pop();

                if (idx == levelSize - 1) {
                    result.push_back(node->val);
                }

                if (node->left) {
                    nq.push(node->left);
                }
                if (node->right) {
                    nq.push(node->right);
                }
            }
        }

        return result;
    }
};