给定一个二叉搜索树和一个目标节点K。任务是找到具有给定目标值K的绝对差最小的节点。
例子：
// For above binary search tree
Input : k = 4
Output : 4
Input : k = 18
Output : 17
Input : k = 12
Output : 9
解决此问题的简单方法是将给定二叉搜索树的有序遍历存储在辅助数组中，然后通过取每个元素的绝对差来找到在线性时间内具有给定目标值K的最小绝对差的节点。
一个有效的解决方案是利用BST的特性。这是解决此问题的算法：
- 如果目标值K存在于给定的BST中，则它是具有最小绝对差的节点。
- 如果目标值K小于当前节点的值，则移至左子级。
- 如果目标值K大于当前节点的值，则移至右子级。
C++
// Recursive C++ program to find key closest to k // in given Binary Search Tree. #includeusing namespace std; /* A binary tree node has key, pointer to left child and a pointer to right child */ struct Node { int key; struct Node* left, *right; }; /* Utility that allocates a new node with the given key and NULL left and right pointers. */ struct Node* newnode(int key) { struct Node* node = new (struct Node); node->key = key; node->left = node->right = NULL; return (node); } // Function to find node with minimum absolute // difference with given K // min_diff --> minimum difference till now // min_diff_key --> node having minimum absolute // difference with K void maxDiffUtil(struct Node *ptr, int k, int &min_diff, int &min_diff_key) { if (ptr == NULL) return ; // If k itself is present if (ptr->key == k) { min_diff_key = k; return; } // update min_diff and min_diff_key by checking // current node value if (min_diff > abs(ptr->key - k)) { min_diff = abs(ptr->key - k); min_diff_key = ptr->key; } // if k is less than ptr->key then move in // left subtree else in right subtree if (k < ptr->key) maxDiffUtil(ptr->left, k, min_diff, min_diff_key); else maxDiffUtil(ptr->right, k, min_diff, min_diff_key); } // Wrapper over maxDiffUtil() int maxDiff(Node *root, int k) { // Initialize minimum difference int min_diff = INT_MAX, min_diff_key = -1; // Find value of min_diff_key (Closest key // in tree with k) maxDiffUtil(root, k, min_diff, min_diff_key); return min_diff_key; } // Driver program to run the case int main() { struct Node *root = newnode(9); root->left = newnode(4); root->right = newnode(17); root->left->left = newnode(3); root->left->right = newnode(6); root->left->right->left = newnode(5); root->left->right->right = newnode(7); root->right->right = newnode(22); root->right->right->left = newnode(20); int k = 18; cout << maxDiff(root, k); return 0; }
Java
// Recursive Java program to find key closest to k // in given Binary Search Tree. class solution { static int min_diff, min_diff_key; /* A binary tree node has key, pointer to left child and a pointer to right child */ static class Node { int key; Node left, right; }; /* Utility that allocates a new node with the given key and null left and right pointers. */ static Node newnode(int key) { Node node = new Node(); node.key = key; node.left = node.right = null; return (node); } // Function to find node with minimum absolute // difference with given K // min_diff -. minimum difference till now // min_diff_key -. node having minimum absolute // difference with K static void maxDiffUtil(Node ptr, int k) { if (ptr == null) return ; // If k itself is present if (ptr.key == k) { min_diff_key = k; return; } // update min_diff and min_diff_key by checking // current node value if (min_diff > Math.abs(ptr.key - k)) { min_diff = Math.abs(ptr.key - k); min_diff_key = ptr.key; } // if k is less than ptr.key then move in // left subtree else in right subtree if (k < ptr.key) maxDiffUtil(ptr.left, k); else maxDiffUtil(ptr.right, k); } // Wrapper over maxDiffUtil() static int maxDiff(Node root, int k) { // Initialize minimum difference min_diff = 999999999; min_diff_key = -1; // Find value of min_diff_key (Closest key // in tree with k) maxDiffUtil(root, k); return min_diff_key; } // Driver program to run the case public static void main(String args[]) { Node root = newnode(9); root.left = newnode(4); root.right = newnode(17); root.left.left = newnode(3); root.left.right = newnode(6); root.left.right.left = newnode(5); root.left.right.right = newnode(7); root.right.right = newnode(22); root.right.right.left = newnode(20); int k = 18; System.out.println( maxDiff(root, k)); } } //contributed by Arnab Kundu
Python3
# Recursive Python program to find key # closest to k in given Binary Search Tree. # Utility that allocates a new node with the # given key and NULL left and right pointers. class newnode: # Constructor to create a new node def __init__(self, data): self.key = data self.left = None self.right = None # Function to find node with minimum # absolute difference with given K # min_diff --> minimum difference till now # min_diff_key --> node having minimum absolute # difference with K def maxDiffUtil(ptr, k, min_diff, min_diff_key): if ptr == None: return # If k itself is present if ptr.key == k: min_diff_key[0] = k return # update min_diff and min_diff_key by # checking current node value if min_diff > abs(ptr.key - k): min_diff = abs(ptr.key - k) min_diff_key[0] = ptr.key # if k is less than ptr->key then move # in left subtree else in right subtree if k < ptr.key: maxDiffUtil(ptr.left, k, min_diff, min_diff_key) else: maxDiffUtil(ptr.right, k, min_diff, min_diff_key) # Wrapper over maxDiffUtil() def maxDiff(root, k): # Initialize minimum difference min_diff, min_diff_key = 999999999999, [-1] # Find value of min_diff_key (Closest # key in tree with k) maxDiffUtil(root, k, min_diff, min_diff_key) return min_diff_key[0] # Driver Code if __name__ == '__main__': root = newnode(9) root.left = newnode(4) root.right = newnode(17) root.left.left = newnode(3) root.left.right = newnode(6) root.left.right.left = newnode(5) root.left.right.right = newnode(7) root.right.right = newnode(22) root.right.right.left = newnode(20) k = 18 print(maxDiff(root, k)) # This code is contributed by PranchalK
C#
using System; // Recursive C# program to find key closest to k // in given Binary Search Tree. public class solution { public static int min_diff, min_diff_key; /* A binary tree node has key, pointer to left child and a pointer to right child */ public class Node { public int key; public Node left, right; } /* Utility that allocates a new node with the given key and null left and right pointers. */ public static Node newnode(int key) { Node node = new Node(); node.key = key; node.left = node.right = null; return (node); } // Function to find node with minimum absolute // difference with given K // min_diff -. minimum difference till now // min_diff_key -. node having minimum absolute // difference with K public static void maxDiffUtil(Node ptr, int k) { if (ptr == null) { return; } // If k itself is present if (ptr.key == k) { min_diff_key = k; return; } // update min_diff and min_diff_key by checking // current node value if (min_diff > Math.Abs(ptr.key - k)) { min_diff = Math.Abs(ptr.key - k); min_diff_key = ptr.key; } // if k is less than ptr.key then move in // left subtree else in right subtree if (k < ptr.key) { maxDiffUtil(ptr.left, k); } else { maxDiffUtil(ptr.right, k); } } // Wrapper over maxDiffUtil() public static int maxDiff(Node root, int k) { // Initialize minimum difference min_diff = 999999999; min_diff_key = -1; // Find value of min_diff_key (Closest key // in tree with k) maxDiffUtil(root, k); return min_diff_key; } // Driver program to run the case public static void Main(string[] args) { Node root = newnode(9); root.left = newnode(4); root.right = newnode(17); root.left.left = newnode(3); root.left.right = newnode(6); root.left.right.left = newnode(5); root.left.right.right = newnode(7); root.right.right = newnode(22); root.right.right.left = newnode(20); int k = 18; Console.WriteLine(maxDiff(root, k)); } } // This code is contributed by Shrikant13
输出：
17时间复杂度： O(h)其中h是给定二叉搜索树的高度。