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

我的解答：
首先通过中序遍历将所有节点的值都存放在一个vector<int> a中，然后遍历树，对每个节点加上a中比节点要大的值
这样写十分臃肿，时间复杂度也高。

/** * 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);    }};

比较好的做法是直接利用中序遍历，同时，在类中声明一个全局的sum变量，初始值为0。
利用迭代的思想
对于一个结点，他本身需要加上右子树中所有节点的和，以及父节点已经改变了的值。

/** * 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);    }};