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 538: https://leetcode.com/problems/convert-bst-to-greater-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
[1, 100]. 0 <= Node.val <= 100- All the values in the tree are unique.
rootis guaranteed to be a valid binary search tree.
这道题让我们将一棵二叉搜索树 Binary Search Tree 转为一棵较大树，跟之前那道题 Convert BST to Greater Tree 一模一样，博主不知道出这道题的意义是什么，连背景不换的，LeetCode 为啥还允许完全一样的题目存在，难道是强行帮我们复习之前的题目么？不管了，权当是复习吧。题目中说变为新的较大树的方法是，将 BST 中的每个结点值都加上所有大于其的结点值，由于 BST 的特点是左子结点值小于根结点值，小于右子结点值，则整个 BST 中最大的结点值应该在最右子结点，由于没有再比它更大的了，所以它不用再加上其他的结点值。其根结点值是第二大的结点值，需要加上该右子结点值。该根结点的左子结点（存在的话）就是第三大的结点值，需要加上更新后的根结点值。眼尖的童鞋应该已经看出来了，这个顺序其实就是颠倒后的中序遍历，正常的中序遍历是左根右，这里是右根左，不过没关系，用递归还是一样的简单。这里用一个变量 cur 来记录当前的结点值之和，作为递归函数的一个引用参数。在递归函数中，先判空，然后对右子结点调用递归函数，之后将 cur 值加到当前结点上面，然后更新 cur 值为当前结点值，然后再对左子结点调用递归函数即可，参见代码如下：
class Solution {
public:
TreeNode* bstToGst(TreeNode* root) {
int cur = 0;
helper(root, cur);
return root;
}
void helper(TreeNode*& node, int& cur) {
if (!node) return;
helper(node->right, cur);
node->val += cur;
cur = node->val;
helper(node->left, cur);
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1038
类似题目：
参考资料：
https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/