Binary-Tree

Given a non-empty special binary tree consisting of nodes with the non-negative value, where each node in this tree has exactly two or zero sub-node. If the node has two sub-nodes, then this node’s value is the smaller value among its two sub-nodes. More formally, the property root.val = min(root.left.val, root.right.val) always holds. Given such a binary tree, you need to output the second minimum value in the set made of all the nodes’ value in the whole tree. ...

Given the root of a binary tree, return all root-to-leaf paths in any order. A leaf is a node with no children. Example 1: Input: root = [1,2,3,null,5] Output: ["1->2->5","1->3"] Example 2: Input: root = [1] Output: ["1"] Constraints: The number of nodes in the tree is in the range [1, 100]. -100 <= Node.val <= 100 Solution /** * 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<String> binaryTreePaths(TreeNode root) { List<String> res = new ArrayList<>(); dfs(root, res, ""); return res; } void dfs(TreeNode node, List<String> res, String s) { if (node == null) { return; } if (s.length() == 0) { s += "" + node.val; } else { s += "->" + node.val; } if (node.left == null && node.right == null) { res.add(s); return; } dfs(node.left, res, s); dfs(node.right, res, s); } }

Given the root of a binary search tree, return a balanced binary search tree with the same node values. If there is more than one answer, return any of them. A binary search tree is balanced if the depth of the two subtrees of every node never differs by more than 1. Example 1: Input: root = [1,null,2,null,3,null,4,null,null] Output: [2,1,3,null,null,null,4] Explanation: This is not the only correct answer, [3,1,4,null,2] is also correct. ...

Given two binary search trees root1 and root2. Return a list containing all the integers from both trees sorted in ascending order. Example 1: Input: root1 = [2,1,4], root2 = [1,0,3] Output: [0,1,1,2,3,4] Example 2: Input: root1 = [0,-10,10], root2 = [5,1,7,0,2] Output: [-10,0,0,1,2,5,7,10] Example 3: Input: root1 = [], root2 = [5,1,7,0,2] Output: [0,1,2,5,7] Example 4: Input: root1 = [0,-10,10], root2 = [] Output: [-10,0,10] Example 5: Input: root1 = [1,null,8], root2 = [8,1] Output: [1,1,8,8] ...

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST): BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST. boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false. int next() Moves the pointer to the right, then returns the number at the pointer. Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST. ...

314. Binary Tree Vertical Order Traversal Given the root of a binary tree, return the vertical order traversal of its nodes’ values. (i.e., from top to bottom, column by column). If two nodes are in the same row and column, the order should be from left to right. Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[9],[3,15],[20],[7]] Example 2: Input: root = [3,9,8,4,0,1,7] Output: [[4],[9],[3,0,1],[8],[7]] Example 3: Input: root = [3,9,8,4,0,1,7,null,null,null,2,5] Output: [[4],[9,5],[3,0,1],[8,2],[7]] ...

987. Vertical Order Traversal of a Binary Tree Given the root of a binary tree, calculate the vertical order traversal of the binary tree. For each node at position (row, col), its left and right children will be at positions (row + 1, col - 1) and (row + 1, col + 1) respectively. The root of the tree is at (0, 0). The vertical order traversal of a binary tree is a list of top-to-bottom orderings for each column index starting from the leftmost column and ending on the rightmost column. There may be multiple nodes in the same row and same column. In such a case, sort these nodes by their values. ...

515. Find Largest Value in Each Tree Row Given the root of a binary tree, return an array of the largest value in each row of the tree (0-indexed). Example 1: Input: root = [1,3,2,5,3,null,9] Output: [1,3,9] Example 2: Input: root = [1,2,3] Output: [1,3] Example 3: Input: root = [1] Output: [1] Example 4: Input: root = [1,null,2] Output: [1,2] Example 5: Input: root = [] Output: [] ...

144. Binary Tree Preorder Traversal 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] Example 2: Input: root = [] Output: [] Example 3: Input: root = [1] Output: [1] Example 4: Input: root = [1,2] Output: [1,2] Example 5: Input: root = [1,null,2] Output: [1,2] ...

108. Convert Sorted Array to Binary Search Tree Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree. A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one. Example 1: Input: nums = [-10,-3,0,5,9] Output: [0,-3,9,-10,null,5] Explanation: [0,-10,5,null,-3,null,9] is also accepted: ...