Trim A Binary Search Tree Problem
Description
LeetCode Problem 669.
Given the root of a binary search tree and the lowest and highest boundaries as low and high, trim the tree so that all its elements lies in [low, high]. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node’s descendant should remain a descendant). It can be proven that there is a unique answer.
Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.
Example 1:
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Input: root = [1,0,2], low = 1, high = 2
Output: [1,null,2]
Example 2:
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Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
Output: [3,2,null,1]
Example 3:
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Input: root = [1], low = 1, high = 2
Output: [1]
Example 4:
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Input: root = [1,null,2], low = 1, high = 3
Output: [1,null,2]
Example 5:
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Input: root = [1,null,2], low = 2, high = 4
Output: [2]
Constraints:
- The number of nodes in the tree in the range [1, 10^4].
- 0 <= Node.val <= 10^4
- The value of each node in the tree is unique.
- root is guaranteed to be a valid binary search tree.
- 0 <= low <= high <= 10^4
Sample C++ Code
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* trimBST(TreeNode* root, int L, int R) {
if (root == NULL)
return NULL;
TreeNode* node;
if (root->val < L) {
node = root->right;
root->right = NULL;
return trimBST(node, L, R);
} else if (root->val > R) {
node = root->left;
root->left = NULL;
return trimBST(node, L, R);
} else {
root->left = trimBST(root->left, L, R);
root->right = trimBST(root->right, L, R);
return root;
}
}
};