二叉树的右视图

给定一棵二叉树，想象自己站在它的右侧，按照从顶部到底部的顺序，返回从右侧所能看到的节点值。

示例:

输入: [1,2,3,null,5,null,4]
输出: [1, 3, 4]
解释:

1 <---
/
2 3 <---

5 4 <---

code：bfs

/**
 * 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) {
        if(root==nullptr)
            return {};
        
        vector<int> res;
        queue<TreeNode*> q;
        q.push(root);
        TreeNode* last=root,*nlast=nullptr;
        while(!q.empty())
        {
            root=q.front();
            q.pop();
            if(root->left)
            {
                q.push(root->left);
                nlast=root->left;
            }
            if(root->right)
            {
                q.push(root->right);
                nlast=root->right;
            }
            if(root==last)
            {
                res.push_back(root->val);
                last=nlast;
            }
        }
        return res;
    }
};

 code：递归，根，右，左，当当前深度比最大深度大时，放入结果集

/**
 * 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 {
private:
    void rightSideViewCore(TreeNode* root,vector<int>& res,int deep,int& maxDeep)
    {
        if(root==nullptr)
            return ;

        if(deep>maxDeep)
        {
            maxDeep=deep;
            res.push_back(root->val);
        }
        rightSideViewCore(root->right,res,deep+1,maxDeep);
        rightSideViewCore(root->left,res,deep+1,maxDeep);
    }
public:
    vector<int> rightSideView(TreeNode* root) {
        if(root==nullptr)
            return {};
        
        vector<int> res;
        int maxDeep=0;
        rightSideViewCore(root,res,1,maxDeep);
        return res;
    }
};