Author: devangini123

Problem Statement: Given the root of a binary tree, invert the tree, and return its root. Examples: Example 1: Input: root = [4,2,7,1,3,6,9] Output: [4,7,2,9,6,3,1] Example 2: Input: root = [2,1,3]] Output: [2,3,1] Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 100]. -100 <= Node.val <= 100 Approach Base Case: If the current node is null, return null. Swap the left and right child of the current node. Recurse on the left and right subtrees. Finally, return the root node (which now has inverted subtrees). Time Complexity:…

Problem Statement: Given the root of a binary tree, check whether it is a mirror of itself(i.e., symmetric around its center). Examples: Example 1: Input: root = [1,2,2,3,4,4,3] Output: true Example 2: Input: root = [1,2,2,null,3,null,3] Output: false Constraints: The number of nodes in the tree is in the range [1, 1000]. -100

Problem Statement: Given the root of a binary tree, check whether it is a mirror of itself(i.e., symmetric around its center). Examples: Example 1: Input: root = [1,2,2,3,4,4,3] Output: true Example 2: Input: root = [1,2,2,null,3,null,3] Output: false Constraints: The number of nodes in the tree is in the range [1, 1000]. -100 val == right->val) && isMirror(left->left, right->right) && isMirror(left->right, right->left); } bool isSymmetric(TreeNode* root) { return isMirror(root->left, root->right); } bool isMirror(struct TreeNode* left, struct TreeNode* right) { if (!left && !right) return true; if (!left || !right) return false; return (left->val == right->val) && isMirror(left->left, right->right) && isMirror(left->right,…

Problem Statement: Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum. A leaf is a node with no children. Examples: Example 1: Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22 Output: true Explanation: The root-to-leaf path with the target sum is shown. Example 2: Input: root = [1,2,3], targetSum = 5 Output: false Explanation: There are two root-to-leaf paths in the tree: (1 –> 2): The sum is 3. (1 –> 3): The sum is 4. There is…

Problem Statement: Given the root of a binary tree, return its maximum depth. A binary tree’s maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Examples: Example 1: Input: root = [3,9,20,null,null,15,7] Output: 3 Example 2: Input: root = [1,null,2] Output: 2 Constraints: The number of nodes in the tree is in the range [0, 104]. -100 left), maxDepth(root->right)) + 1; } #include <stdio.h> #include <stdlib.h> struct TreeNode { int val; struct TreeNode *left; struct TreeNode *right; }; int max(int a, int b) { return a > b…

Problem Statement: Given the root of a binary tree, return the level order traversal of its nodes’ values.. (i.e., from left to right, level by level). Examples: Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]] Example 2: Input: root = [1] Output: [[1]] Explanation: Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 100]. -100

Problem Statement: Given the root of a binary tree, return the level order traversal of its nodes’ values.. (i.e., from left to right, level by level). Examples: Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]] Example 2: Input: root = [1] Output: [[1]] Explanation: Example 3: Input: root = [] Output: [] Constraints: The number of nodes in the tree is in the range [0, 100]. -100 <= Node.val <= 100 Approach Base Case: If the tree is empty (root == null), return an empty array. Initialize Queue: Use a queue to process nodes level by level. Start by pushing…

DFS and BFS in Binary Tree: DFS: Depth First Search DFS explores a tree by going as deep as possible along a branch before backtracking. BFS: Breadth First Search Explore the tree level by level, explore all nodes at the current level before moving deeper. DFS vs BFS DFS: Preorder: A B D E C F Inorder: D B E A C F Postorder: D E B F C A BFS: Level Order Traversal: A B C D E F Zig-Zag Traversal: Zig-Zag Traversal: A C B D E F DFS / BFS DFS uses a Stack (LIFO) First Explore…

🌙 Problem Statement: Given the root of a binary tree, return the postorder traversal of its nodes’ values.. Examples: Example 1: Input: root = [1,null,2,3] Output: [3,2,1] Explanation: Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] Output: [4,6,7,5,2,9,8,3,1] Explanation: Example 3: Input: root = [] Output: [] Example 4: Input: root = [1] Output: [1] Constraints: The number of nodes in the tree is in the range [0, 100]. -100 right && peek->right != lastVisited) { curr = peek->right; } else { ans.push_back(peek->val); lastVisited = stack.top(); stack.pop(); } } return ans; } #include <stdio.h> #include <stdlib.h> #include <stdbool.h> #define MAX 1000 typedef…

🌙 Problem Statement: Given the root of a binary tree, return the postorder traversal of its nodes’ values.. Examples: Example 1: Input: root = [1,null,2,3] Output: [3,2,1] Explanation: Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] Output: [4,6,7,5,2,9,8,3,1] Explanation: Example 3: Input: root = [] Output: [] Example 4: Input: root = [1] Output: [1] Constraints: The number of nodes in the tree is in the range [0, 100]. -100 left); if (curr->right) s1.push(curr->right); } vector ans; while (!s2.empty()) { ans.push_back(s2.top()->val); s2.pop(); } return ans; } void postorderTraversal(struct TreeNode* root, int* result, int* returnSize) { if (!root) return; struct TreeNode* s1[1000]; struct…