[LeetCode] 270. Closest Binary Search Tree Value 最近的二分搜索樹的值

 

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.

Example:

Input: root = [4,2,5,1,3], target = 3.714286

    4
   / \
  2   5
 / \
1   3

Output: 4

 

這道題讓我們找一個二分搜索數的跟給定值最接近的一個節點值，由於是二分搜索樹，所以博主最先想到用中序遍歷來做，一個一個的比較，維護一個最小值，不停的更新，實際上這種方法並沒有提高效率，用其他的遍歷方法也可以，參見代碼如下：

 

解法一：

class Solution {
public:
    int closestValue(TreeNode* root, double target) {
        double d = numeric_limits<double>::max();
        int res = 0;
        stack<TreeNode*> s;
        TreeNode *p = root;
        while (p || !s.empty()) {
            while (p) {
                s.push(p);
                p = p->left;
            }
            p = s.top(); s.pop();
            if (d >= abs(target - p->val)) {
                d = abs(target - p->val);
                res = p->val;
            }
            p = p->right;
        }
        return res;
    }
};

 

實際我們可以利用二分搜索樹的特點 (左<根<右) 來快速定位，由於根節點是中間值，在往下遍歷時，根據目標值和根節點的值大小關系來比較，如果目標值小於節點值，則應該找更小的值，於是到左子樹去找，反之去右子樹找，參見代碼如下：

 

解法二：

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;
    }
};

 

以上兩種方法都是迭代的方法，下面來看遞歸的寫法，下面這種遞歸的寫法和上面迭代的方法思路相同，都是根據二分搜索樹的性質來優化查找，但是遞歸的寫法用的是回溯法，先遍歷到葉節點，然后一層一層的往回走，把最小值一層一層的運回來，參見代碼如下：

 

解法三：

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;
    }
};

 

再來看另一種遞歸的寫法，思路和上面的都相同，寫法上略有不同，用if來分情況，參見代碼如下：

 

解法三：

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;
    }
};

 

最后來看一種分治法的寫法，這種方法相當於解法一的遞歸寫法，並沒有利用到二分搜索樹的性質來優化搜索，參見代碼如下：

 

解法四：

class Solution {
public:
    int closestValue(TreeNode* root, double target) {
        double diff = numeric_limits<double>::max();
        int res = 0;
        helper(root, target, diff, res);
        return res;
    }
    void helper(TreeNode *root, double target, double &diff, int &res) {
        if (!root) return;
        if (diff >= abs(root->val - target)) {
            diff = abs(root->val - target);
            res = root->val;
        }
        helper(root->left, target, diff, res);
        helper(root->right, target, diff, res);
    }
};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/270

 

類似題目：

Count Complete Tree Nodes

Closest Binary Search Tree Value II

Search in a Binary Search Tree

 

參考資料:

https://leetcode.com/problems/closest-binary-search-tree-value/

https://leetcode.com/problems/closest-binary-search-tree-value/discuss/70331/Clean-and-concise-java-solution

https://leetcode.com/problems/closest-binary-search-tree-value/discuss/70322/Super-clean-recursive-Java-solution

https://leetcode.com/problems/closest-binary-search-tree-value/discuss/70327/4-7-lines-recursiveiterative-RubyC%2B%2BJavaPython

 

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