英文:
Understanding the solution to LeetCode question 1038 - Convert BST to greater tree
问题
在我为LeetCode问题1038找到的解决方案中，代码如下所示：
const bstToGst = (root) => {
function go(root, sum) {
if (!root) return sum;
root.val += go(root.right, sum);
return go(root.left, root.val);
}
go(root, 0);
return root;
};
我理解这段代码中的if (!root) return sum和root.val += go(root.right, sum)这两行的目的，但我对return go(root.left, root.val)语句的目的理解有困难。能有人详细解释一下这段代码在做什么吗？
顺便说一下，你可以使用以下示例根节点（如果你想用示例来解释）：
4
/ \
1 6
/ \ / \
0 2 5 7
\ \
3 8
英文:
In the solution I found for LeetCode question 1038, the code looks like this:
const bstToGst = (root) => {
function go(root, sum) {
if (!root) return sum;
root.val += go(root.right, sum);
return go(root.left, root.val);
}
go(root, 0);
return root;
};
I understand the purpose of the lines if (!root) return sum and root.val += go(root.right, sum), but I'm having trouble understanding the reason for the return go(root.left, root.val) statement. Can someone explain what this code is doing in more detail?
By the way, as example root you can use (if you want to explain with an example):
4
/ \
1 6
/ \ / \
0 2 5 7
\ \
3 8
答案1
得分: 2
返回的语句 go(root.left, root.val); 用于继续遍历到当前节点的左子节点，将当前节点的更新值作为新的总和传递。这是基于二叉搜索树的属性，其中左子树中的所有节点都小于或等于当前节点。因此，这个总和将添加到左子树中的所有节点。这个过程递归地继续，直到树中的所有节点都被更新。
该解决方案基本上使用了二叉搜索树的属性和中序遍历（在这种情况下是右根左顺序）来在单次遍历中高效解决问题。
英文:
The return go(root.left, root.val); statement is used to continue the traversal to theleft child of the current node, passing the updated value of the current node as the new sum. This is based on the property of the binary search tree where all nodes in the left subtree are less than or equal to the current node. Therefore, this sum will be added to all the nodes in the left subtree. This process continues recursively until all nodes in the tree are updated.
The solution basically uses the properties of Binary Search Tree and in-order traversal (right-root-left order in this case) to solve the problem efficiently in a single pass.
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