538. Convert BST to Greater Tree

題目描述

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

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

解析

1.我的方法
先中序遍歷得到遞增序列，求得大於每個結點的結點的val之和，然後再通過DFS算法將其加到每個結點的val上。比較麻煩=.=

class Solution {
private:
    vector<int> v;
    unordered_map<int,int> mapp;
public:
    void inOrder(TreeNode* root){
        if(root==NULL) return;
        inOrder(root->left);
        v.push_back(root->val);
        inOrder(root->right);
    }
    void dfs(TreeNode* root){
        if(root==NULL) return;
        root->val+=mapp[root->val];
        dfs(root->left);
        dfs(root->right);
    }
    TreeNode* convertBST(TreeNode* root) {
        inOrder(root);
        for(int i=v.size()-1;i>=0;i--){
            if(i!=v.size()-1) mapp[v[i]]=v[i+1]+mapp[v[i+1]];
        }
        dfs(root);
        return root;
    }
};

2.回溯法
思想：遍歷一個沒有遍歷過的節點之前，先將大於點值的點都遍歷一遍。用到的是反序中序遍歷 。
遞歸算法：首先我們判斷當前訪問的節點是否存在，如果存在就遞歸右子樹，遞歸回來的時候更新總和和當前點的值，然後遞歸左子樹。

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