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11 changes: 11 additions & 0 deletions week06/LCA_of_a_binary_tree.cpp
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// Time complexity: O(N) , N = Number of nodes.
// Space complexity: O(1).
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Space complexity is O(H), where H = max depth of the tree. It is because of the stack memory used by the recursion.


TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q){
if(!root) return NULL;
if(root->val == p->val || root->val == q->val) return root;
TreeNode* leftNode = lowestCommonAncestor(root->left, p, q);
TreeNode* rightNode = lowestCommonAncestor(root->right, p, q);
if(leftNode && rightNode) return root;
return leftNode ? leftNode : rightNode;
}
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// Time complexity: O(N), N = Number of Nodes.
// Space complexity: O(N), N = Number of Nodes.

TreeNode* treeBuilding(vector<int> &preorder, vector<int> &inorder, int left, int right, int &pos, unordered_map<int, int> &pathMap){
if(left > right) return NULL;
TreeNode* currNode = new TreeNode(preorder[pos++]);
int mid = pathMap[currNode->val];
currNode->left = treeBuilding(preorder, inorder, left, mid - 1, pos, pathMap);
currNode->right = treeBuilding(preorder, inorder, mid + 1, right, pos, pathMap);
return currNode;
}

TreeNode* buildTree(vector<int> &preorder, vector<int> &inorder){
int pos = 0;
unordered_map<int, int> pathMap;
int arrLength = inorder.size();
for(int i = 0; i < arrLength; i++){
pathMap[inorder[i]] = i;
}
TreeNode* root = treeBuilding(preorder, inorder, 0, preorder.size() - 1, pos, pathMap);
return root;
}
17 changes: 17 additions & 0 deletions week06/convert_sorted_array_to_binary_search_tree.cpp
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// Time complexity: O(N), N = Number of nodes;
// Space complexity: O(N), N = Number of nodes;

TreeNode* buildBST(vector<int>& nums, int left, int right){
if(left > right) return NULL;

int mid = left + (right - left) / 2;
TreeNode* currNode = new TreeNode(nums[mid]);
currNode->left = buildBST(nums, left, mid - 1);
currNode->right = buildBST(nums, mid + 1, right);

return currNode;
}

TreeNode* sortedArray(vector<int>& nums){
return buildBST(nums, 0, nums.size() - 1);
}
19 changes: 19 additions & 0 deletions week06/kth_smallest_element_in_a_bst.cpp
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// Time complexity: O(N), N = Number of nodes;
// Space complexity: O(1)


void traverseTree(TreeNode* root, int &k, int &kthValue){
if(!root) return;
traverseTree(root->left, k, kthValue);
k--;
if(k == 0){
kthValue = root->val;
}
traverseTree(root->right, k, kthValue);
}

int kthSmallest(TreeNode* root, int k){
int kthValue;
traverseTree(root, k, kthValue);
return kthValue;
}
7 changes: 7 additions & 0 deletions week06/maximum_depth_of_binary_tree.cpp
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// Time complexity: O(N), N = Number of nodes
// Space complexity: O(1)
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Space complexity is O(H), where H = max depth of the tree. It is because of the stack memory used by the recursion.


int maxDepth(TreeNode* root, int depth = 0){
if(!root) return depth;
return max(maxDepth(root->left, depth + 1), maxDepth(root->right, depth + 1));
}
16 changes: 16 additions & 0 deletions week06/populating_next_right_pointers_in_each_node.cpp
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// Time complexity: O(N), N = Number of nodes
// Space complexity: O(1)
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Space complexity is O(H), where H = max depth of the tree. It is because of the stack memory used by the recursion.


void traverseNode(Node* node){
if(!root) return;
if(node->left) node->left->next = node->right;
if(node->right && node->next) node->right->next = node->next->left;

if(node->left) traverseNode(node->left);
if(node->right) traverseNode(node->right);
}

Node* connect(Node* root){
traverseNode(root);
return root;
}
31 changes: 31 additions & 0 deletions week06/symmetric_tree.cpp
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// Time complexity: O(N), N = Number of nodes;
// Space complexity: O(N), N = Number of nodes;

void nodeTraverseLeft(TreeNode* node, vector<int> &u){
if(!node){
u.push_back(1000); // i chose a random value outside of range.
return;
}
u.push_back(node->val);
nodeTraversalLeft(node->left, u);
nodeTraversalLeft(node->right, u);
}
void nodeTraverseRight(TreeNode* node, vector<int> &v){
if(!node){
v.push_back(1000); // i chose a random value outside of range.
return;
}
v.push_back(node->val);
nodeTraversalLeft(node->left, v);
nodeTraversalLeft(node->right, v);
}

bool isSymmetric(TreeNode* root){
vector<int> a, b;
nodeTraverseLeft(root->left, a);
nodeTraverseRight(root->right, a);
for(int i = 0; i < a.size(); i++){
if(a[i] != b[i]) return false;
}
return true;
}
15 changes: 15 additions & 0 deletions week06/validate_binary_search_tree.cpp
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// Time complexity: O(N), N = Number of nodes;
// Space complexity: O(1)
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Space complexity is O(H), where H = max depth of the tree. It is because of the stack memory used by the recursion.


bool isBST(TreeNode* node, long long left, long long right){
if(!root) return true;
long long node_value = node->val;
if(node_value >= left && node_value <= right){
return isBST(node->left, left, node_value - 1) && isBST(node->right, node_value + 1, right);
}
return false;
}

bool isValidBST(TreeNode* root){
return isBST(root, LLONG_MIN, LLONG_MAX);
}