把二叉搜索树变成一个二叉树，使得原来每个节点的值 变成 (原来的值+所有大于它的值的和)

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

Examples:
Input: Root of following BST
                5
              /   \
           2     13
Output: The given BST is converted to following Binary Tree
              18
              /   \
          20     13

使用递归的方法，递归函数定义为 convert(Node* root, int& sum)，root是子树的根节点，sum是该子树所有节点的和。

方法一代码如下：

#include<stdio.h>#include<stdlib.h> struct Node{          int value;  Node* lchild;  Node* rchild;          Node(int val, Node* lch, Node* rch):value(val), lchild(lch), rchild(rch) {}       };void convert(Node* root, int& sum){if(root==NULL)return;convert(root->rchild, sum);root->value = sum + root->value;sum = root->value;convert(root->lchild, sum);}void print(Node* root){if(root){printf("%d|",root->value);print(root->lchild);  print(root->rchild);}}int main(){    Node* n1 = new Node(1,NULL,NULL);      Node* n3 = new Node(3,NULL,NULL);      Node* n2 = new Node(2,n1,n3);      Node* n5 = new Node(5,NULL,NULL);      Node* n4 = new Node(4,n2,n5);   /* Constructed binary tree is             4           /   \         2      5       /  \     1     3   */  int sum = 0;   convert(n4, sum);// convert(n4, 15) is wrong, since a temporary object cannot be assigned to a non-const referenceprint(n4);getchar();return 0;}

方法二也采用递归：

#include<stdio.h>#include<stdlib.h> struct Node{          int value;  Node* lchild;  Node* rchild;          Node(int val, Node* lch, Node* rch):value(val), lchild(lch), rchild(rch) {}       };void convert(Node* root, int& sum){if(root==NULL)return;if(root->lchild)convert(root->lchild, sum);int temp = root->value;root->value = sum;sum = sum - temp;convert(root->rchild, sum);}void print(Node* root){if(root){printf("%d|",root->value);print(root->lchild);  print(root->rchild);}}int main(){    Node* n1 = new Node(1,NULL,NULL);      Node* n3 = new Node(3,NULL,NULL);      Node* n2 = new Node(2,n1,n3);      Node* n5 = new Node(5,NULL,NULL);      Node* n4 = new Node(4,n2,n5);   /* Constructed binary tree is             4           /   \         2      5       /  \     1     3   */  int sum = 15;   convert(n4, sum);// convert(n4, 15) is wrong, since a temporary object cannot be assigned to a non-const referenceprint(n4);getchar();return 0;}