[LeetCode] 270. Closest Binary Search Tree Value 最近的二叉搜索树的值

Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.

Note:

• Given target value is a floating point.
• You are guaranteed to have only one unique value in the BST that is closest to the target.

给一个非空二叉树和一个目标值，找到和目标值最接近的一个节点值。

利用二分搜索树的特点(左<根<右)来快速定位，由于根节点是中间值，在遍历时，如果目标值小于节点值，则找更小的值到左子树去找，反之去右子树找。

解法1：迭代

解法2：递归

Java：

public int closestValue(TreeNode root, double target) {
    double min=Double.MAX_VALUE;
    int result = root.val;
 
    while(root!=null){
        if(target>root.val){
 
            double diff = Math.abs(root.val-target);
            if(diff<min){
                min = Math.min(min, diff);
                result = root.val;
            }
            root = root.right;
        }else if(target<root.val){
 
            double diff = Math.abs(root.val-target);
            if(diff<min){
                min = Math.min(min, diff);
                result = root.val;
            }
            root = root.left;
        }else{
            return root.val;
        }
    }
 
    return result;
}　　

Java:

public class Solution {
    int goal;
    double min = Double.MAX_VALUE;
 
    public int closestValue(TreeNode root, double target) {
        helper(root, target);
        return goal;
    }
 
    public void helper(TreeNode root, double target){
        if(root==null)
            return;
 
        if(Math.abs(root.val - target) < min){
            min = Math.abs(root.val-target);
            goal = root.val;
        } 
 
        if(target < root.val){
            helper(root.left, target);
        }else{
            helper(root.right, target);
        }
    }
}　

Java: Iteration

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int closestValue(TreeNode root, double target) {
        if (root == null) return 0;
        int min = root.val;
        while (root != null) {
            min = (Math.abs(root.val - target) < Math.abs(min - target) ? root.val : min);
            root = (root.val < target) ? root.right : root.left;
        }
        return min;
    }
}

Java: Recursion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int closestValue(TreeNode root, double target) {
        TreeNode child = target < root.val ? root.left : root.right;
        if (child == null) {
            return root.val; 
        }
        int childClosest = closestValue(child, target);
        return Math.abs(root.val - target) < Math.abs(childClosest - target) ? root.val : childClosest;
    }
}

Python: Iteration, Time: O(h), Space: O(1)

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def closestValue(self, root, target):
        """
        :type root: TreeNode
        :type target: float
        :rtype: int
        """
        gap = float("inf")
        closest = float("inf")
        while root:
            if abs(root.val - target) < gap:
                gap = abs(root.val - target)
                closest = root
            if target == root.val:
                break
            elif target < root.val:
                root = root.left
            else:
                root = root.right
        return closest.val

C++: Iteration

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int closestValue(TreeNode* root, double target) {
        double gap = numeric_limits<double>::max();
        int closest = numeric_limits<int>::max();
        
        while (root) {
            if (abs(static_cast<double>(root->val) - target) < gap) {
                gap = abs(root->val - target);
                closest = root->val;
            }
            if (target == root->val) {
                break;
            } else if (target < root->val) {
                root = root->left;
            } else {
                root = root->right;
            }
        }
        return closest;
    }
};　

C++: Iteration

class Solution {
public:
    int closestValue(TreeNode* root, double target) {
        int res = root->val;
        while (root) {
            if (abs(res - target) >= abs(root->val - target)) {
                res = root->val;
            }
            root = target < root->val ? root->left : root->right;
        }
        return res;
    }
};

C++: Recursion

class Solution {
public:
    int closestValue(TreeNode* root, double target) {
        int a = root->val;
        TreeNode *t = target < a ? root->left : root->right;
        if (!t) return a;
        int b = closestValue(t, target);
        return abs(a - target) < abs(b - target) ? a : b;
    }
};

C++: Recursion

class Solution {
public:
    int closestValue(TreeNode* root, double target) {
        int res = root->val;
        if (target < root->val && root->left) {
            int l = closestValue(root->left, target);
            if (abs(res - target) >= abs(l - target)) res = l;
        } else if (target > root->val && root->right) {
            int r = closestValue(root->right, target);
            if (abs(res - target) >= abs(r - target)) res = r;
        }
        return res;
    }
};　　

类似题目：

　　

All LeetCode Questions List 题目汇总