104. Maximum Depth of Binary Tree

Solution - Recursion

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root)
    {
        if(!root) return 0;
        return 1 + max(maxDepth(root->left), maxDepth(root->right));
    }
};

• T:
  $O(N)$
• S:
  $O(\log N)$

Solution - Iteration

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root)
    {
        if (!root) return 0;
        queue<TreeNode*> q{{root}};
        int depth = 0;
        while (!q.empty())
        {
            ++depth;
            int qSize = q.size();
            for (int i = 0; i < qSize; ++i)
            {
                TreeNode* node = q.front(); q.pop();
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return depth;
    }
};

• T:
  $O(N)$
• S:
  $O(\log N)$