Maximum Depth of Binary Tree

• LeetCode: Maximum Depth of Binary Tree
• Problem:

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

• Analysis:

The leaf node feature is that left and right children are null.

• Code - Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        if (root.left == null && root.right == null) {
            return 1;
        }
        return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
    }
}

• Code - C++:

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

• Code - Python:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root == None:
            return 0
        if root.left == None and root.right == None:
            return 1
        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

• Code - C:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int maxDepth(struct TreeNode* root) {
    if (root == 0) {
        return 0;
    }
    if (root->left == 0 && root->right == 0) {
        return 1;
    }
    int left = maxDepth(root->left);
    int right = maxDepth(root->right);
    int val = left < right? right : left;
    return 1 + val;
}

• Time Complexity: O(N), N is the number of nodes.
• Space Complexity: In-memory