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
解题思路：按照题目意思，是从最右右子树开始累加sum，所以可以按照右子树 -> 中间节点 -> 左子树 这样的顺序去对这个二叉查找树进行累加节点数值。
1 #include <stdio.h> 2 3 struct TreeNode { 4 int val; 5 TreeNode *left; 6 TreeNode *right; 7 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 }; 9 10 void travel_tree(TreeNode *node, int &sum){ 11 if (!node){ 12 return; 13 } 14 travel_tree(node->right, sum); 15 sum += node->val; 16 node->val = sum; 17 travel_tree(node->left, sum); 18 } 19 20 class Solution { 21 public: 22 TreeNode* convertBST(TreeNode *root) { 23 int sum = 0; 24 travel_tree(root, sum); 25 return root; 26 } 27 }; 28 29 void preorder_print(TreeNode *node, int layer){ 30 if (!node){ 31 return; 32 } 33 for (int i = 0; i < layer; i++){ 34 printf("-----"); 35 } 36 printf("[%d]\n", node->val); 37 preorder_print(node->left, layer + 1); 38 preorder_print(node->right, layer + 1); 39 } 40 41 int main(){ 42 TreeNode a(5); 43 TreeNode b(3); 44 TreeNode c(6); 45 TreeNode d(2); 46 TreeNode e(4); 47 TreeNode f(7); 48 a.left = &b; 49 a.right = &c; 50 b.left = &d; 51 b.right = &e; 52 c.right = &f; 53 Solution solve; 54 TreeNode *root = solve.convertBST(&a); 55 preorder_print(&a, 0); 56 return 0; 57 }
