Populating Next Right Pointers in Each Node - LeetCode You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The…leetcode.com var connect = function(root) { let temp = root; while (temp && temp.left) { const next = temp.left; while (temp) { temp.left.next = temp.right; temp.right.next = temp.next && temp.next.left; temp = temp.next } temp = next; } return root; };

Odd Even Linked List - LeetCode Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even…leetcode.com var oddEvenList = function(head) { if (head === null || head.next === null) return head; let odd = head; let even = head.next; let evenhead = even; while (odd.next !== null && even.next !== null) { odd.next = even.next odd = odd.next even.next = odd.next even = even.next }

Random Pick with Weight - LeetCode You are given a 0-indexed array of positive integers w where w[i] describes the weight of the index. You need to…leetcode.com /** * @param {number[]} w */ var Solution = function(w) { this.arr = []; let sum = 0; for(let i = 0; i < w.length; i++) { sum += w[i]; this.arr.push(sum); } }; /** * @return {number} */ Solution.prototype.pickIndex = function() { let target = Math.random() * this.arr[this.arr.length-1]; let low = 0; let high = this.arr.length - 1; while (low < high) { let mid = Math.floor((low + high)/2) if (this.arr[mid] <= target) low = mid + 1; else high = mid; } return high; };

Path Sum III - LeetCode Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along…leetcode.com var pathSum = function(root, sum) { if (!root) return 0 return ( pathSumStart(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum) ) }; const pathSumStart = (root, sum) => { if (!root) return 0; const self = root.val === sum ? 1 : 0; return ( self + pathSumStart(root.left, sum - root.val) + pathSumStart(root.right, sum - root.val) ) }

Step-By-Step Directions From a Binary Tree Node to Another - LeetCode You are given the root of a binary tree with n nodes. Each node is uniquely assigned a value from 1 to n. You are also…leetcode.com var getDirections = function(root, startValue, destValue) { const startPath = []; const destPath = []; find(root, startValue, startPath); find(root, destValue, destPath); // find the root while (startPath.length && destPath.length && startPath.at(-1) === destPath.at(-1)) { startPath.pop(); destPath.pop(); } // L -> U return startPath.map(() => "U").join("") + destPath.reverse().join(""); };

Binary Tree Zigzag Level Order Traversal - LeetCode Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. (i.e., from left to…leetcode.com function zigzagLevelOrder(root) { let res = []; go(root, 0, res); return res; } function go(node, l, res) { // l means level if (!node) return; if (!res[l]) { res.push([]); } if (l % 2 === 0) { res[l].push(node.val); } else { res[l].unshift(node.val); } go(node.left, l+1, res); go(node.right, l+1, res); }

Longest Univalue Path - LeetCode Given the root of a binary tree, return the length of the longest path, where each node in the path has the same value…leetcode.com var longestUnivaluePath = function(root) { if (!root) return 0; return Math.max( longestPassPath(root.left, root.val) + longestPassPath(root.right, root.val), longestUnivaluePath(root.left), longestUnivaluePath(root.right) ) };

Verify Preorder Serialization of a Binary Tree - LeetCode One way to serialize a binary tree is to use preorder traversal. When we encounter a non-null node, we record the…leetcode.com var isValidSerialization = function(preorder) { const stack = [0]; const arr = preorder.split(','); for(let item of arr) { if(item === '#') stack[stack.length-1]++; else stack.push(0); while(stack.length > 1 && stack[stack.length-1] === 2) { stack.pop(); stack[stack.length-1]++ } } return stack.length === 1 && stack.pop() === 1; };