538. Convert BST to Greater Tree
2017-03-19 16:08
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Given 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.
Example:
题解:
这题就是说对每个节点,加上所有比它大的值。
说简单也简单,关键在于二分查找树。
利用二分查找树的特性,我们可以很容易得出从大到小的数组序列,即中序遍历(右子树-根节点-左子树),如例子中给的就是【13,5,2】。
那么就很明显了,一次遍历,加上前面的数,如5变为5+13=18, 2变为2+18=20。
总体复杂度O(n)。
Example:
Input: The root of a Binary Search Tree like this: 5 / \ 2 13 Output: The root of a Greater Tree like this: 18 / \ 20 13
题解:
这题就是说对每个节点,加上所有比它大的值。
说简单也简单,关键在于二分查找树。
利用二分查找树的特性,我们可以很容易得出从大到小的数组序列,即中序遍历(右子树-根节点-左子树),如例子中给的就是【13,5,2】。
那么就很明显了,一次遍历,加上前面的数,如5变为5+13=18, 2变为2+18=20。
总体复杂度O(n)。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: void convert(TreeNode* root, int& sum){ if(root == NULL) return; convert(root->right, sum); root->val += sum; sum = root->val; convert(root->left, sum); } TreeNode* convertBST(TreeNode* root) { int sum = 0; convert(root, sum); return root; } };
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