[LeetCode] 538. Convert BST to Greater Tree 将二叉搜索树BST转为较大树

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, 104].
-104 <= Node.val <= 104
All the values in the tree are unique.
root is guaranteed to be a valid binary search tree.

这道题让我们将二叉搜索树转为较大树，通过题目汇总的例子可以明白，是把每个结点值加上所有比它大的结点值总和当作新的结点值。仔细观察题目中的例子可以发现，2变成了 20，而 20 是所有结点之和，因为2是最小结点值，要加上其他所有结点值，所以肯定就是所有结点值之和。5变成了 18，是通过 20 减去2得来的，而 13 还是 13，是由 20 减去7得来的，而7是2和5之和。博主开始想的方法是先求出所有结点值之和，然后开始中序遍历数组，同时用变量 sum 来记录累加和，根据上面分析的规律来更新所有的数组。但是通过看论坛，发现还有更巧妙的方法，不用先求出的所有的结点值之和，而是巧妙的将中序遍历左根右的顺序逆过来，变成右根左的顺序，这样就可以反向计算累加和 sum，同时更新结点值，叼的不行，参见代码如下：

解法一：

class Solution {
public:
    TreeNode* convertBST(TreeNode* root) {
        int sum = 0;
        helper(root, sum);
        return root;
    }
    void helper(TreeNode*& node, int& sum) {
        if (!node) return;
        helper(node->right, sum);
        node->val += sum;
        sum = node->val;
        helper(node->left, sum);
    }
};

下面这种方法写的更加简洁一些，没有写其他递归函数，而是把自身写成了可以递归调用的函数，参见代码如下：

解法二：

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

private:
    int sum = 0;
};

下面这种解法是迭代的写法，因为中序遍历有递归和迭代两种写法，逆中序遍历同样也可以写成迭代的形式，参加代码如下：

解法三：

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