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270. Closest Binary Search Tree Value

Posted on Edited on In Disqus:
Symbols count in article: 1.6k Reading time ≈ 1 mins.

iterative O(h) time O(1) space

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/**
 * 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) {
        int res = root->val;
        while (root) {
            if (fabs(root->val - target) < fabs(res - target)) {
                res = root->val;
            }
            root = root->val > target ? root->left : root->right;
        }
        return res;
    }
};

recursive O(h) time O(1) space

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/**
 * 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) {
        res = root->val;
        dfs(root, target);
        return res;
    }

    void dfs(TreeNode *root, double target) {
        if (!root) return;
        if (fabs(root->val - target) < fabs(res - target)) {
            res = root->val;
        }
        root->val > target ? dfs(root->left, target) : dfs(root->right, target);
    }

    int res;
};

O(n) time O(h) space

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/**
 * 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) {
        res = root->val;
        dfs(root, target);
        return res;
    }

    void dfs(TreeNode *root, double target) {
        if (!root) return;
        if (fabs(root->val - target) < fabs(res - target)) {
            res = root->val;
        }
        dfs(root->left, target);
        dfs(root->right, target);
    }

    int res;
};