面试必考精华版Leetcode199. 二叉树的右视图

题目：

---

代码（首刷看解析）：

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        unordered_map<int,int> rightmostvalue;
        queue nodeQueue;
        queue<int> depthQueue;
        int maxDepth=-1;
 
        nodeQueue.push(root);
        depthQueue.push(0);
        while(!nodeQueue.empty()){
            TreeNode* node = nodeQueue.front();nodeQueue.pop();
            int depth = depthQueue.front();depthQueue.pop();
            
 
            if(node!=nullptr){
                maxDepth = max(maxDepth,depth);
                //下面这一行运用了unordered_map的特性更新，直到最右值
                rightmostvalue[depth]=node->val;
 
                nodeQueue.push(node->left);
                nodeQueue.push(node->right);
                depthQueue.push(depth+1);
                depthQueue.push(depth+1);
            }
        }
 
        vector<int> ans;
        for(int i=0;i<=maxDepth;++i){
            ans.push_back(rightmostvalue[i]);
        }
        return ans;
    }
};

代码（二刷自解 2024年1月24日）

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> res = {};
        if(root == nullptr) return res;
        queue que;
        TreeNode* cur = root;
        que.push(cur);
        int size = 0;
        while (!que.empty()) {
            size = que.size();
            while (size--){
                TreeNode* node  = que.front();
                if (size == 0) {
                    res.emplace_back(node->val);
                }
                que.pop();
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return res;
    }
};

---

代码(三刷自解 2024年3月6日）

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> res;
        if (!root) return res;
        queue q;
        TreeNode* cur = root;
        q.push(cur);
        while (!q.empty()) {
            int size = q.size();
            while (size--) {
                TreeNode* node = q.front();
                if (size == 0) res.emplace_back(node->val);
                q.pop();
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return res;
    }
};