树08

108.将有序数组转换为二叉搜索树

/**
 * 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) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        int n = nums.size();
        if (n == 0) return nullptr;
        int half = n / 2;
        TreeNode* root = new TreeNode(nums[half]);
        vector<int> leftnums(nums.begin(), nums.begin() + half);
        vector<int> rightnums(nums.begin() + half + 1, nums.end());
        root -> left = sortedArrayToBST(leftnums);
        root -> right = sortedArrayToBST(rightnums);
        return root;
    }
};

538.把二叉搜索树转换为累加树

/**
 * 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) {}
 * };
 */
class Solution {
public:
    int count = 0;
    TreeNode* convertBST(TreeNode* root) {
        if (root == nullptr) return nullptr;
        convertBST(root -> right);
        root -> val = root -> val + count;
        count = root -> val;
        convertBST(root -> left);
        return root;
    }
};