701. Insert into a Binary Search Tree
- we just have to traverse in the tree till we encounter a leaf.
- while traversing we must obey the rules of BST.
naive kind of implementation
/**
* 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) {}
* };
*/
class Solution {
public:
void insert(TreeNode root, int n) {
}
void dfs(TreeNode* root, int n) {
if (root == nullptr) return;
if (n < root -> val) {
if (root -> left == nullptr) {
root -> left = new TreeNode;
root -> left -> left = nullptr;
root -> left -> right = nullptr;
root -> left -> val = n;
}
else {
dfs(root -> left, n);
}
}
else {
if (root -> right == nullptr) {
root -> right = new TreeNode;
root -> right -> left = nullptr;
root -> right -> right = nullptr;
root -> right -> val = n;
}
else {
dfs(root -> right, n);
}
}
}
TreeNode* insertIntoBST(TreeNode* root, int val) {
if (root == nullptr) {
root = new TreeNode;
root -> left = nullptr;
root -> right = nullptr;
root -> val = val;
return root;
}
dfs(root, val);
return root;
}
};
more concise Solution
TreeNode* insertIntoBST(TreeNode *node, int val) {
if (!node) {
TreeNode *newNode = new TreeNode(val);
return newNode;
}
if (val < node->val) {
node->left = insertIntoBST(node->left, val);
}
else {
node->right = insertIntoBST(node->right, val);
}
return node;
}