leetcode每日一题-99. Recover Binary Search Tree

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
const int N=1010;
class Solution {
public:
    int a[1010],idx=0;
    int cnt=0;//记录dfs遍历结点是第几个
    void getList(TreeNode* u){
        if(u==nullptr) return;
        
        getList(u->left);
        a[idx++]=u->val;
        getList(u->right);
    }
    void changeVal(TreeNode* u){
        if(u==nullptr) return;
        
        changeVal(u->left);
        if(u->val!=a[cnt]) u->val=a[cnt];
        cnt++;
        changeVal(u->right);
    }
    //中序遍历一次得到序列，sort然后重新中序修改
    TreeNode* recoverTree(TreeNode* root) {
        memset(a,0,sizeof a);
        getList(root);
        sort(a,a+idx);
        changeVal(root);
        return root;
    }
};