Hard
Closest Nodes Queries in a Binary Search Tree — C++
Full explanation · Time O(n + qlogn) · Space O(n)
// Time: O(n + qlogn)
// Space: O(n)
// iterative dfs, binary search
class Solution {
public:
vector<vector<int>> closestNodes(TreeNode* root, vector<int>& queries) {
const auto& iter_dfs = [&]() {
vector<int> inorder;
vector<pair<int, TreeNode *>> stk = {{1, root}};
while (!empty(stk)) {
const auto [step, node] = stk.back(); stk.pop_back();
if (step == 1) {
if (!node) {
continue;
}
stk.emplace_back(1, node->right);
stk.emplace_back(2, node);
stk.emplace_back(1, node->left);
} else if (step == 2) {
inorder.emplace_back(node->val);
}
}
return inorder;
};
const auto& inorder = iter_dfs();
vector<vector<int>> result;
for (const auto& q : queries) {
const auto cit = lower_bound(cbegin(inorder), cend(inorder), q);
if (cit == cend(inorder)) {
result.push_back({*prev(cit), -1});
} else if (*cit == q) {
result.push_back({*cit, *cit});
} else if (cit == cbegin(inorder)) {
result.push_back({-1, *cit});
} else {
result.push_back({*prev(cit), *cit});
}
}
return result;
}
};
// Time: O(n + qlogn)
// Space: O(n)
// dfs, binary search
class Solution2 {
public:
vector<vector<int>> closestNodes(TreeNode* root, vector<int>& queries) {
vector<int> inorder;
const function<void (TreeNode*)> dfs = [&](TreeNode *node) {
if (!node) {
return;
}
dfs(node->left);
inorder.emplace_back(node->val);
dfs(node->right);
};
dfs(root);
vector<vector<int>> result;
for (const auto& q : queries) {
const auto cit = lower_bound(cbegin(inorder), cend(inorder), q);
if (cit == cend(inorder)) {
result.push_back({*prev(cit), -1});
} else if (*cit == q) {
result.push_back({*cit, *cit});
} else if (cit == cbegin(inorder)) {
result.push_back({-1, *cit});
} else {
result.push_back({*prev(cit), *cit});
}
}
return result;
}
};