题解 | #输出二叉树的右视图#

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     * 求二叉树的右视图
     * @param xianxu int整型vector 先序遍历
     * @param zhongxu int整型vector 中序遍历
     * @return int整型vector
     */
    //重构二叉树
    TreeNode* reConstruct(vector<int> pre, vector<int> vin)
    {
        if(pre.size() == 0) return nullptr;
        TreeNode* head = new TreeNode(pre[0]);
        vector<int> preL, vinL, preR, vinR;
        int len = vin.size();
        int gen = 0;
        for(int i = 0; i < len; ++i)
        {
            if(vin[i] == pre[0])
            {
                gen = i;
                break;
            }
        }
        for(int i = 0; i < gen; ++i)
        {
            preL.push_back(pre[i+1]);
            vinL.push_back(vin[i]);
        }
        for(int i = gen+1; i < len; ++i)
        {
            preR.push_back(pre[i]);
            vinR.push_back(vin[i]);
        }
        head->left = reConstruct(preL, vinL);
        head->right = reConstruct(preR, vinR);
        return head;
    }
    //层序遍历，非递归实现，每一层的最后一个节点即是右视图
    vector<int> solve(vector<int>& xianxu, vector<int>& zhongxu) {
        // write code here
        TreeNode* head = reConstruct(xianxu, zhongxu);
        vector<int> rst;
        if(head == nullptr) return rst;
        queue<TreeNode* > que;
        que.push(head);
        while(!que.empty())
        {
            int size = que.size();
            while(size--)
            {
                TreeNode* node = que.front();
                que.pop();
                if(size == 0) rst.push_back(node->val);
                if(node->left) que.push(node->left);
                if(node->right) que.push(node->right);
            }
        }
        return rst;
    }
};