Binary Search Tree     Binary Tree     C++     Depth-First Search     Medium     Tree    

Problem Statement:

Given the root of 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 the sum of all keys greater than the original key in BST.

As a reminder, a binary search tree is a tree that satisfies these constraints:

  • The left subtree of a node contains only nodes with keys less than the node's key.
  • The right subtree of a node contains only nodes with keys greater than the node's key.
  • Both the left and right subtrees must also be binary search trees.

 

Example 1:

Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Example 2:

Input: root = [0,null,1]
Output: [1,null,1]

 

Constraints:

  • The number of nodes in the tree is in the range [1, 100].
  • 0 <= Node.val <= 100
  • All the values in the tree are unique.

 

Note: This question is the same as 538: https://leetcode.com/problems/convert-bst-to-greater-tree/

Solution:

Given that all keys in the tree are unique we can use a HashMap of the structure unordered_map<int,TreeNode*> to store the tree. Then we can do the operations asked and substitute appropriate values as the new keys.

 
class Solution {
public:
    void traverse(TreeNode *root, unordered_map<int, TreeNode *>&H)
    {
        if (root==NULL) return;
        H[root->val] = root;
        traverse(root->left, H);
        traverse(root->right, H);
    }
    TreeNode* bstToGst(TreeNode* root) {
        unordered_map<int, TreeNode *> H;
        traverse(root, H);
        vector<int> v1, v2;
        for (auto item: H) v1.push_back(item.first);
        sort(v1.begin(),v1.end(), greater<int>());
        int curr=0;
        for (int i: v1)
        {
            v2.push_back(curr+i);
            curr+=i;
        }
        for (int i=0; i<v1.size(); i++) H[v1[i]]->val = v2[i];
        return root;
    }
};