CodeWithClarity

Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. Note: A leaf is a node with no children.   Example 1: Input: root = [3,9,20,null,null,15,7] Output: 2 Example 2: Input: root = [2,null,3,null,4,null,5,null,6] Output: 5   Constraints: The number of nodes in the tree is in the range [0, 105]. -1000 <= Node.val <= 1000 /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *         this.right = right;  *     }  * }  *...

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.   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 <= Node.val <= 100  /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *         this.right = right;  *     }  * }  */ class Solution {   ...

 Given the root of a binary tree, return the sum of all left leaves. A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.   Example 1: Input: root = [3,9,20,null,null,15,7] Output: 24 Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively. Example 2: Input: root = [1] Output: 0   Constraints: The number of nodes in the tree is in the range [1, 1000]. -1000 <= Node.val <= 1000 /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *         this.right = right; ...

 Given the roots of two binary trees p and q, write a function to check if they are the same or not. Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.   Example 1: Input: p = [1,2,3], q = [1,2,3] Output: true Example 2: Input: p = [1,2], q = [1,null,2] Output: false Example 3: Input: p = [1,2,1], q = [1,1,2] Output: false   Constraints: The number of nodes in both trees is in the range [0, 100]. -104 <= Node.val <= 104 /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *        ...

 Given the root of a complete binary tree, return the number of the nodes in the tree. According to Wikipedia, every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h. Design an algorithm that runs in less than O(n) time complexity.   Example 1: Input: root = [1,2,3,4,5,6] Output: 6 Example 2: Input: root = [] Output: 0 Example 3: Input: root = [1] Output: 1   Constraints: The number of nodes in the tree is in the range [0, 5 * 104]. 0 <= Node.val <= 5 * 104 The tree is guaranteed to be complete.  Brute Force Approach: DFS (O(n)) Just do any traversal and count nodes. /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *    ...

 Given the root of a binary tree, return the postorder traversal of its nodes' values.   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 the nodes in the tree is in the range [0, 100]. -100 <= Node.val <= 100   Follow up: Recursive solution is trivial, could you do it iteratively?  Steps (Recursive Approach) Traverse left subtree Traverse right subtree Process current node (root) /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) { ...

 Given the root of a binary tree, return the inorder traversal of its nodes' values.   Example 1: Input: root = [1,null,2,3] Output: [1,3,2] Explanation: Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] Output: [4,2,6,5,7,1,3,9,8] Explanation: Example 3: Input: root = [] Output: [] Example 4: Input: root = [1] Output: [1] Approach 1: Recursive (Briefly) /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *         this.right = right;  *     }  * }  */ class Solution {     public List < Integer > i...

 Given the root of a binary tree, return the preorder traversal of its nodes' values.   Example 1: Input: root = [1,null,2,3] Output: [1,2,3] Explanation: Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] Output: [1,2,4,5,6,7,3,8,9] Explanation: Example 3: Input: root = [] Output: [] Example 4: Input: root = [1] Output: [1]  Approach 1: Recursive (Simple) 🔹 Steps: Visit the current node → add value to result Recursively traverse the left subtree Recursively traverse the right subtree /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *  ...

 You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST. Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.   Example 1: Input: root = [4,2,7,1,3], val = 5 Output: [4,2,7,1,3,5] Explanation: Another accepted tree is: Example 2: Input: root = [40,20,60,10,30,50,70], val = 25 Output: [40,20,60,10,30,50,70,null,null,25] Example 3: Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5 Output: [4,2,7,1,3,5]    Recursive Approach (Simple & Clean) 🔹 Steps: If root is null , create a new node with the given value. If val < root.val , go to the left subtree . If val > root.val , go to the right subtree . At the correct position (null child), insert a new node . Return the root . /**...

 You are given the root of a binary search tree (BST) and an integer val. Find the node in the BST that the node's value equals val and return the subtree rooted with that node. If such a node does not exist, return null.   Example 1: Input: root = [4,2,7,1,3], val = 2 Output: [2,1,3] Example 2: Input: root = [4,2,7,1,3], val = 5 Output: [] /**  * Definition for a binary tree node.  * public class TreeNode {  *     int val;  *     TreeNode left;  *     TreeNode right;  *     TreeNode() {}  *     TreeNode(int val) { this.val = val; }  *     TreeNode(int val, TreeNode left, TreeNode right) {  *         this.val = val;  *         this.left = left;  *         this.right = right;  *     }  * }  */ class Solution {     public TreeNode searchBST ( TreeNode root , int val ...