199. Binary Tree Right Side View**

​​https://leetcode.com/problems/binary-tree-right-side-view/​​

题目描述

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.

Example:

Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

C++ 实现 1

层序遍历, 先存入右孩子即可.

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

C++ 实现 2

另外这题使用 DFS 也可以完成, 在 ​​dfs​​​ 中先遍历右子树, 再遍历左子树. 使用 ​​max_depth​​​ 记录当前访问的最大深度, 如果 ​​depth > max_depth​​​ 时, 对 ​​max_depth​​​ 和 ​​res​​ 进行更新.

class Solution {
private:
    vector<int> res;
    int max_depth = 0;
    void dfs(TreeNode *root, int depth) {
        if (!root) return;
        if (depth > max_depth) {
            max_depth = depth;
            res.push_back(root->val);
        }
        dfs(root->right, depth + 1);
        dfs(root->left, depth + 1);
    }
public:
    vector<int> rightSideView(TreeNode* root) {
        if (!root) return {};
        res.push_back(root->val);
        dfs(root, 0);
        return res;
    }
};