LeetCode OJ：Maximum Depth of Binary Tree（二叉树最大深度）

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.

求二叉树深度，直接看代码：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int maxDepth(TreeNode* root) {
13         if(root == NULL) return 0;
14         _max = 1;
15         tranverse(root->left, 1);
16         tranverse(root->right, 1);
17         return _max;
18     }
19     void tranverse(TreeNode * node, int count)
20     {
21         if(node == NULL)
22             return;
23         else{ 
24             if(count + 1 > _max)
25                 _max = count + 1;
26             tranverse(node->left, count + 1);
27             tranverse(node->right, count + 1);
28         }
29     }
30 public:
31     int _max;
32 };

java版的如下所示，上面的递归写的还是太麻烦了，下面的简单很多哈：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int maxDepth(TreeNode root) {
12         if(root == null)
13             return 0;
14         return tranverse(root, 0);
15     }
16     
17     public int tranverse(TreeNode root, int val){
18         if(root.left == null && root.right == null)
19             return val+1;
20         int leftDep = 0, rightDep = 0;
21         if(root.left != null)
22             leftDep = tranverse(root.left, val + 1);
23         if(root.right != null)
24             rightDep = tranverse(root.right, val + 1);
25        return Math.max(leftDep, rightDep);
26     }
27 }