538. Convert BST to Greater Tree

Link: https://leetcode.com/problems/convert-bst-to-greater-tree/

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 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.

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

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]

Example 3:

Input: root = [1,0,2]
Output: [3,3,2]

Example 4:

Input: root = [3,2,4,1]
Output: [7,9,4,10]

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題目翻譯:

將二元樹變成 greater 二元樹。

程式思路:

簡單說就是 當前節點等於右邊子樹的總和。
先用postorder 走一遍存入陣列，然後計算後再走一次 postorder。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> arr;
    void postorder(TreeNode *node)
    {
        if(node == NULL)
            return;
        postorder(node->right);
        arr.push_back(node->val);
        postorder(node->left);
    }

    void greater_postorder(TreeNode *node,int &idx)
    {
        if(node == NULL)
            return;
        greater_postorder(node->right,idx);
        node->val = arr[idx++];
        greater_postorder(node->left,idx);
    }

    TreeNode* convertBST(TreeNode* root) {
        postorder(root);

        int sz = arr.size();
        for(int i = 1;i < sz;i++)
        {
            arr[i] = arr[i-1] + arr[i];
        }
        int idx = 0;
        greater_postorder(root,idx);
        return  root;
    }
};