[Leetcode] Trim a Binary Search Tree(Medium)
LeetCode 669 - Trim a Binary Search Tree
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
Input: root = [1,0,2], low = 1, high = 2
Output: [1,null,2]
Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
Output: [3,2,null,1]
How can we solve this problem?
這題就是要我們將一顆BST的少於low的部分以及大於heigh的部分移除。這題打算使用遞歸來解決。我們只要將比low小的Node的右子樹接到他的父節點，並取代比low還小的Node，而比heigh大的Node的左子樹接到他的父節點，並取代比heigh還大的Node即可。
Solution:
/**
* 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 low, int high) {
//space : O(tree node size for all nodes value are between low and height)
//time : O(tree node size for all nodes)
return TrimBST(root,low,high);
}
TreeNode* TrimBST(TreeNode* root,int low,int height){
if(!root) return nullptr;
//if the root value is less than height ,go right sub-tree
//if the root value is greater than low ,go left sub-tree
if(root->val < low){
// root->left = nullptr;
return TrimBST(root->right,low,height);
}else if(root->val > height){
// root->right = nullptr;
return TrimBST(root->left,low,height);
}
//left
root->left = TrimBST(root->left,low,height);
//right
root->right = TrimBST(root->right,low,height);
// cout << root->val << "\n";
return root;
}
};