Tree
Depth-First Search
Binary Search Tree
Binary Tree
Description
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:
Input: root = [1,0,2], low = 1, high = 2
Output: [1,null,2]
Example 2:
Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
Output: [3,2,null,1]
Constraints:
The number of nodes in the tree is in the range [1, 104 ]
.
0 <= Node.val <= 104
The value of each node in the tree is unique .
root
is guaranteed to be a valid binary search tree.
0 <= low <= high <= 104
Solutions
Solution 1
Python3 Java C++ Go TypeScript Rust JavaScript C
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22 # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution :
def trimBST (
self , root : Optional [ TreeNode ], low : int , high : int
) -> Optional [ TreeNode ]:
def dfs ( root ):
if root is None :
return root
if root . val > high :
return dfs ( root . left )
if root . val < low :
return dfs ( root . right )
root . left = dfs ( root . left )
root . right = dfs ( root . right )
return root
return dfs ( root )
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31 /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode trimBST ( TreeNode root , int low , int high ) {
if ( root == null ) {
return root ;
}
if ( root . val > high ) {
return trimBST ( root . left , low , high );
}
if ( root . val < low ) {
return trimBST ( root . right , low , high );
}
root . left = trimBST ( root . left , low , high );
root . right = trimBST ( root . right , low , high );
return root ;
}
}
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22 /**
* 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 ) {
if ( ! root ) return root ;
if ( root -> val > high ) return trimBST ( root -> left , low , high );
if ( root -> val < low ) return trimBST ( root -> right , low , high );
root -> left = trimBST ( root -> left , low , high );
root -> right = trimBST ( root -> right , low , high );
return root ;
}
};
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22 /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func trimBST ( root * TreeNode , low int , high int ) * TreeNode {
if root == nil {
return root
}
if root . Val > high {
return trimBST ( root . Left , low , high )
}
if root . Val < low {
return trimBST ( root . Right , low , high )
}
root . Left = trimBST ( root . Left , low , high )
root . Right = trimBST ( root . Right , low , high )
return root
}
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29 /**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function trimBST ( root : TreeNode | null , low : number , high : number ) : TreeNode | null {
const dfs