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538. Convert BST to Greater Tree

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작성일 2022.04.15 18:25

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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 [0, 104].
-104 <= Node.val <= 104
All the values in the tree are unique.
root is guaranteed to be a valid binary search tree.

Note: This question is the same as 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

#leetcode, #문제풀이, #시즌3, #medium, #amazon, #facebook, #bloomberg, #tree, #depth-first search, #binary search tree, #binary tree

링크

https://leetcode.com/problems/convert-bst-to-greater-tree/ 596 회 연결

익명
작성일 2022.04.16 12:40

C++
Runtime: 36 ms, faster than 90.34% of C++ online submissions for Convert BST to Greater Tree.
Memory Usage: 33.5 MB, less than 62.62% of C++ online submissions for Convert BST to Greater Tree.

class Solution {
private:
    int sum = 0;
    
public:
    TreeNode* convertBST(TreeNode* root) {
        if (root) {
            convertBST(root->right);
            root->val = (sum += root->val);
            convertBST(root->left);
        }
        return root;
    }
};

익명
작성일 2022.04.17 15:34

Runtime: 51 ms, faster than 22.70% of C++ online submissions for Convert BST to Greater Tree.
Memory Usage: 33.5 MB, less than 19.98% of C++ online submissions for Convert BST to Greater Tree.

class Solution {
public:
    
    void sumBST(TreeNode* &p, int &sum){
        
        if(p->right != nullptr)
            sumBST(p->right,sum);
        p->val = sum + p->val;
        
        sum=p->val;
        
        if(p->left != nullptr)
            sumBST(p->left,sum);
    };
    
    TreeNode* convertBST(TreeNode* root) {
        int sum=0;
        if(root!=nullptr){
            sumBST(root,sum);
        };
        return root;
    }
};