DFS

Problem Statement Description Given the root of a binary tree, return the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root. The length of a path between two nodes is represented by the number of edges between them. Examples Example 1: flowchart TD a((1))-->b((2)) b-->c((4)) b-->d((5)) a-->e((3)) Input: root = [1,2,3,4,5] Output: 3 Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3]....

Tree Node Definition # Definition for a binary tree node. from typing import List class TreeNode: def __init__(self, x: int) -> None: self.val = x self.left = None self.right = None Recursive Implementation left ➡️ right ➡️ root class Solution: def traversal(self, root: 'TreeNode', result=None) -> List[List[int]]: """ 0 / \ 1 2 / \ 3 4 / \ 6 9 Post-order...: [3, 6, 9, 4, 1, 2, 0] T: O(N) S: O(N) worst case; O(log N) average """ if result is None: result = [] if not root: return [] self....

Tree Node Definition # Definition for a binary tree node. from typing import List class TreeNode: def __init__(self, x: int) -> None: self.val = x self.left = None self.right = None Recursive Implementation left ➡️ root ➡️ right class Solution: def traversal(self, root: 'TreeNode', result=None) -> List[List[int]]: """ 0 / \ 1 2 / \ 3 4 / \ 6 9 In-order...: [3, 1, 6, 4, 9, 0, 2] T: O(N) S: O(N) worst case; O(log N) average """ if result is None: result = [] if not root: return [] self....

Tree Node Definition # Definition for a binary tree node. from typing import List class TreeNode: def __init__(self, x: int) -> None: self.val = x self.left = None self.right = None Recursive Implementation root ➡️ left ➡️ right class Solution: def traversal(self, root: 'TreeNode', ans=None) -> List[List[int]]: """ 0 / \ 1 2 / \ 3 4 / \ 6 9 Pre-order...: [0, 1, 3, 4, 5, 9, 2] """ if not root: return [] if ans is None: ans = [] ans....

Problem Statement Given a list of accounts where each element accounts[i] is a list of strings, where the first element accounts[i][0] is a name, and the rest of the elements are emails representing emails of the account. Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some common email to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name....