538. Convert BST to Greater Tree

Given a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.

Example:

Input: The root of a Binary Search Tree like this:
              5
            /   \
           2     13

Output: The root of a Greater Tree like this:
             18
            /   \
          20     13

我的解答：
首先通過中序遍歷將所有節點的值都存放在一個vector<int> a中，然後遍歷樹，對每個節點加上a中比節點要大的值
這樣寫十分臃腫，時間複雜度也高。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* convertBST(TreeNode* root) {
        vector<int> a;
        toVector(root, a);
        addToGreater(root, a);
        return root;
    }
    
    void addToGreater(TreeNode* root, vector<int>& a){
        if(!root){
            return;
        }
        int temp = root->val;
        for(int i = 0; i < a.size(); i++){
            if(a[i] > temp){
                root->val += a[i];
            }
        }
        addToGreater(root->left, a);
        addToGreater(root->right, a);
    }
    
    void toVector(TreeNode* root, vector<int>& a){
        if(!root){
            return;
        }
        toVector(root->left, a);
        a.push_back(root->val);
        toVector(root->right, a);
    }
};

比較好的做法是直接利用中序遍歷，同時，在類中聲明一個全局的sum變量，初始值爲0。
利用迭代的思想
對於一個結點，他本身需要加上右子樹中所有節點的和，以及父節點已經改變了的值。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sum = 0;
    TreeNode* convertBST(TreeNode* root) {
        addSum(root);
        return root;
    }
    void addSum(TreeNode* root){
        if(!root){
            return;
        }
        addSum(root->right);
        root->val = (sum += root->val);
        addSum(root->left);
    }
};