二叉搜索树（平衡二叉搜索树）

1、验证是否为二叉搜索树
中序遍历看是否为递增序列

2、平衡二叉搜索树
将升序序列转为二叉平衡搜索树

class Solution {
public:
    TreeNode* dfs(vector& nums, int start, int end){
        if(start > end){
            return nullptr;
        }
        int mid = (start+end)/2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = dfs(nums, start, mid-1);
        root->right = dfs(nums, mid+1, end);
        return root;
    }
    TreeNode* sortedArrayToBST(vector& nums) {
        TreeNode* res = dfs(nums, 0, nums.size()-1);
        return res;
    }
};

思路：将二叉搜索树先变为升序序列（用第一题的方法），再将升序序列变成平衡二叉搜索树（用第二题的方法）

class Solution {
public:
    vector vec;
    void getvalue(TreeNode* root){
        if(root == nullptr){
            return;
        }
        getvalue(root->left);
        vec.push_back(root->val);
        getvalue(root->right);
    }
    TreeNode* makeBST(vector nums, int start, int end){
        if(start > end){
            return nullptr;
        }
        int mid = (start+end)/2;
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = makeBST(nums, start, mid-1);
        root->right = makeBST(nums, mid+1, end);
        return root;
    }
    TreeNode* balanceBST(TreeNode* root) {
        getvalue(root);
        return makeBST(vec, 0, vec.size()-1);
    }
};