diff --git a/MaximumDepthofBinaryTree.java b/MaximumDepthofBinaryTree.java
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+++ b/MaximumDepthofBinaryTree.java
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+/*
+ Author:     King, wangjingui@outlook.com
+ Date:       Dec 25, 2014
+ Problem:    Maximum Depth of Binary Tree
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/maximum-depth-of-binary-tree/
+ Notes:
+ 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.
+
+ Solution: Recursion.
+ */
+/**
+ * Definition for binary tree
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+public class Solution {
+    public int maxDepth(TreeNode root) {
+        if (root == null) return 0;
+        int left = maxDepth(root.left);
+        int right = maxDepth(root.right);
+        return Math.max(left, right) + 1;
+    }
+}
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diff --git a/MinimumDepthofBinaryTree.java b/MinimumDepthofBinaryTree.java
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+++ b/MinimumDepthofBinaryTree.java
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+/*
+ Author:     King, wangjingui@outlook.com
+ Date:       Dec 25, 2014
+ Problem:    Minimum Depth of Binary Tree
+ Difficulty: easy
+ Source:     https://oj.leetcode.com/problems/minimum-depth-of-binary-tree/
+ Notes:
+ Given a binary tree, find its minimum depth.
+ The minimum depth is the number of nodes along the shortest path from the root node 
+ down to the nearest leaf node.
+ 
+ Solution: 1. Recursion. Pay attention to cases when the non-leaf node has only one child.
+           PS. 2. Iteration + Queue. 
+ */
+
+/**
+ * Definition for binary tree
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+public class Solution {
+    public int minDepth(TreeNode root) {
+        if (root == null) return 0;
+        if (root.left == null) return minDepth(root.right) + 1;
+        if (root.right == null) return minDepth(root.left) + 1;
+        return Math.min(minDepth(root.left), minDepth(root.right)) + 1;
+    }
+}
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