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Two Sum IV - Input is BST.cpp
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Two Sum IV - Input is BST.cpp
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
// Approach 1: Using HashSet[Accepted]
// Time Complexity : O(N)
// Space Complexity : O(N)
class Solution {
public:
bool findTarget(TreeNode* root, int k) {
map<int, bool> mp;
return Helper(root, k, mp);
}
bool Helper(TreeNode* root, int k, map<int, bool> &mp){
if(root == NULL)
return false;
if(mp.count(k - root -> val))
return true;
mp[root->val] = 1;
return (Helper(root->left, k, mp) || Helper(root->right, k, mp));
}
};
// Approach 2: Using BFS
// Time Complexity : O(N)
// Space Complexity : O(N)
class Solution {
public:
bool findTarget(TreeNode* root, int k) {
if(root == NULL)
return false;
queue<TreeNode*> q;
map<int, bool> mp;
q.push(root);
while(!q.empty()){
TreeNode* temp = q.front();
q.pop();
if(temp != NULL){
if(mp[k - temp -> val])
return true;
mp[temp->val] = true;
q.push(temp->left);
q.push(temp->right);
}
}
return false;
}
};
// Approach 3: Inorder Traversal
// Time Complexity : O(N)
// Space Complexity : O(N)
class Solution {
public:
bool findTarget(TreeNode* root, int k) {
if(root == NULL)
return false;
vector<int> arr;
inorder(root, arr);
int l = 0, r = arr.size() - 1;
while(l < r){
int sum = arr[l] + arr[r];
if(sum == k)
return true;
else if(sum < k)
++l;
else
--r;
}
return false;
}
void inorder(TreeNode* root, vector<int> &arr){
stack<TreeNode*> st;
while(root != NULL || !st.empty()){
while(root != NULL){
st.push(root);
root = root->left;
}
root = st.top();
st.pop();
arr.push_back(root->val);
root = root -> right;
}
}
};