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This approach uses the recursive traversal of the BST properties. It checks if the current node's value is equal to the target value. If it is, it returns the subtree rooted at the current node. If the target value is less than the current node, it recurses into the left subtree; if greater, it recurses into the right subtree. This method leverages the BST's inherent properties to narrow down the search.
Time Complexity: O(h) where h is the height of the tree, as we traverse from the root to a leaf.
Space Complexity: O(h) due to the recursion stack.
1class TreeNode:
2 def __init__(self, val=0, left=None, right=None):
3 self.val = val
4 self.left = left
5 self.right = right
6
7class Solution:
8 def searchBST(self, root: TreeNode, val: int) -> TreeNode:
9 if not root or root.val == val:
10 return root
11 if val < root.val:
12 return self.searchBST(root.left, val)
13 return self.searchBST(root.right, val)This Python function searchBST follows a recursive pattern to search for the node. If it finds the node with val, it returns that subtree; otherwise traverses left or right based on the value comparison.
In this approach, we use an iterative solution instead of recursion to save stack space. Starting from the root, we use a loop to traverse the tree. At each node, we check if it matches the target value. If it matches, we return the node; if not, we decide to move to the left or right child based on the BST property. The absence of a matching node returns null.
Time Complexity: O(h) where h is tree height.
Space Complexity: O(1) since we're only using a few extra pointers.
1 public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
this.val = val;
this.left = left;
this.right = right;
}
}
public class Solution {
public TreeNode SearchBST(TreeNode root, int val) {
while (root != null && root.val != val) {
root = val < root.val ? root.left : root.right;
}
return root;
}
}The C# while loop checks each tree node iteratively, navigating left or right based on the value comparison, and outputs either the found node or null.