将二叉搜索树变平衡

class Solution {
public:
    vector<int> v;
    void Tree2vec(TreeNode* root){                  // 中序遍历将二叉搜索树按顺序存在vector上
        if(root->left) Tree2vec(root->left); 
        v.push_back(root->val);
        if(root->right) Tree2vec(root->right);
    }
    TreeNode* vec2Tree(int left, int right){       // 根据vector从中间往两边构造平衡二次搜索树
        if(left>right) return NULL;
        int mid=(left+right)/2;
        TreeNode* result=new TreeNode(v[mid]);
        result->left=vec2Tree(left, mid-1);
        result->right=vec2Tree(mid+1, right);
        return result;
    }
    TreeNode* balanceBST(TreeNode* root) {
        Tree2vec(root);
        int mid=(v.size()-1)/2;
        TreeNode* result=new TreeNode(v[mid]);
        result->left=vec2Tree(0, mid-1);
        result->right=vec2Tree(mid+1, v.size()-1);
        return result;
    }
};