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LeetCode 538. Convert BST to Greater Tree

Posted on Edited on In

题目

原题在此

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]

Constraints:

  • The number of nodes in the tree is in the range [0, $10^4$].
  • $-10^4$ <= Node.val <= $10^4$
  • All the values in the tree are unique.
  • root is guaranteed to be a valid binary search tree.

解析

以右中左的顺序遍历BST, 将节点的值累加写入. 右中左的遍历就是把Inorder反过来.

代码

recursive

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/**
 * 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:
    TreeNode* convertBST(TreeNode* root) {
        if(!root) return NULL;
        sum(root, 0);
        return root;
    }
    int sum(TreeNode* root, int large){
        if(!root) return large;
        root->val += sum(root->right, large);
        return sum(root->left, root->val);
    }
};

iterative

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class Solution {
public:
    TreeNode* convertBST(TreeNode* root) {
        TreeNode* tmp = root;
        stack<TreeNode*> s;
        int sum = 0;
        while(tmp || !s.empty()){
            while(tmp){
                s.push(tmp);
                tmp = tmp -> right;
            }
            tmp = s.top();
            s.pop();
            sum += tmp -> val;
            tmp -> val = sum;
            tmp = tmp -> left;                                
        }
        return root;
    }
};