题解 | #牛群的树形结构展开#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
#include <vector>
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
     *
     * 
     * @param root TreeNode类 
     * @return TreeNode类
     */
    vector<int> v;
    void dfs(TreeNode* root)
    {
        if(!root)
            return;

        v.emplace_back(root->val);
        dfs(root->left);
        dfs(root->right);

    }

    TreeNode* flattenTree(TreeNode* root) {
        // write code here
        if(!root)
            return nullptr;
        dfs(root);
        reverse(v.begin(), v.end());
        // 根据先序遍历
        TreeNode* ans = new TreeNode(v.back());
        v.pop_back();

        queue<TreeNode* >q;
        q.push(ans);

        while(!v.empty())
        {
            int len = q.size();
            for(int i=0; !v.empty() && i<len; ++i)
            {
                TreeNode* t = q.front();
                q.pop();
                
                t->left = nullptr;
                t->right = new TreeNode(v.back());
                v.pop_back();

                q.push(t->right);
            }
        }
        
        return ans;
    }
};