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.
rootis 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;
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/538
参考资料：
https://leetcode.com/problems/convert-bst-to-greater-tree/
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/100506/Java-Recursive-O(n)-time
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/100610/c%2B%2B-solution-beats-100
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