Leetcode 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

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。

#include <stdio.h>

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

void travel_tree(TreeNode *node, int &sum){
    if (!node){
        return;
    }
    travel_tree(node->right, sum);
    sum += node->val;
    node->val = sum;
    travel_tree(node->left, sum);
}

class Solution {
public:
    TreeNode* convertBST(TreeNode *root) {
        int sum = 0;
        travel_tree(root, sum);
        return root;
    }
};

void preorder_print(TreeNode *node, int layer){
    if (!node){
        return;
    }
    for (int i = 0; i < layer; i++){
        printf("-----");
    }
    printf("[%d]
", node->val);
    preorder_print(node->left, layer + 1);
    preorder_print(node->right, layer + 1);
}

int main(){
    TreeNode a(5);
    TreeNode b(3);
    TreeNode c(6);
    TreeNode d(2);
    TreeNode e(4);
    TreeNode f(7);
    a.left = &b;
    a.right = &c;
    b.left = &d;
    b.right = &e;
    c.right = &f;
    Solution solve;
    TreeNode *root = solve.convertBST(&a);
    preorder_print(&a, 0);
    return 0;
}