Maximum Depth of Binary Tree - Leetcode 104 - Python 3 Easy Ways [Recursion, BFS, DFS]

Maximum Depth of Binary Tree

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.

Recusive DFS

class Solution:
    def maxDepth(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        return 1 + max(self.maxDepth(root.left),self.maxDepth(root.right))

BFS

class Solution:
    def maxDepth(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        lvl = 0
        q = deque([root])
        while q:
            for i in range(len(q)):
                node = q.popleft()
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            lvl += 1
        return lvl

Iterative DFS

class Solution:
    def maxDepth(self, root: Optional[TreeNode]) -> int:
        s = [[root,1]] #stack
        ans = 0
        
        while s:
            n, d = s.pop()
            
            if n:
                ans = max(ans, d)
                s.append([n.left, d+1])
                s.append([n.right, d+1])
                
        return ans

Explaination :

Comments

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371 . Sum of Two Integers Given two integers a and b , return the sum of the two integers without using the operators + and - . Example 1: Input: a = 1, b = 2 Output: 3 Example 2: Input: a = 2, b = 3 Output: 5 Constraints: -1000 <= a, b <= 1000 Solution : C++ : class Solution { public: int getSum(int a, int b) { if (b==0) return a; int sum = a ^ b; int cr = (unsigned int) (a & b) << 1; return getSum(sum, cr); } }; Java : class Solution { public int getSum(int a, int b) { while(b != 0){ int tmp = (a & b) << 1; a = a ^ b; b = tmp; } return a; } } Explaination :

Leetcode 217. Contains Duplicate. Python (Easiest Approach ✅)

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Leetcode 322. Coin Change. Python (Greedy? vs DP?)

322 . Coin Change You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. Return the fewest number of coins that you need to make up that amount . If that amount of money cannot be made up by any combination of the coins, return -1 . You may assume that you have an infinite number of each kind of coin. Example 1: Input: coins = [1,2,5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1 Example 2: Input: coins = [2], amount = 3 Output: -1 Example 3: Input: coins = [1], amount = 0 Output: 0 Constraints: 1 <= coins.length <= 12 1 <= coins[i] <= 2 31 - 1 0 <= amount <= 10 4 Solution : class Solution: def coinChange(self, coins: List[int], amount: int) -> int: dp = [amount + 1] * (amount + 1) #[0...7] dp[0] = 0 ...

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