題目：

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root. Example: Given a binary tree

          1
         / \
        2   3
       / \     
      4   5    
 

Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Note: The length of path between two nodes is represented by the number of edges between them.

解釋： 求二叉樹中存在的最長的路徑，這個路徑可以不經過根節點。 這道題目可以通過求二叉樹深度的方法來求，在dfs中返回的是二叉樹的深度，但是dfs的過程中更新當前的最長的路徑，一個結點所經過的路徑的長度，等與它左子樹的深度+右子樹的深度（二叉樹的深度其實就是root和最低的結點之間，連線線的個數）。

python程式碼：

# 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 diameterOfBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
 

        self.max_sum=0
        def TreeDepth(root):
            left_depth=0
            right_depth=0
            if root.left:
                left_depth=TreeDepth(root.left)
            if root.right:
                right_depth=TreeDepth(root.right)
            self.max_sum=max(self.max_sum,left_depth+right_depth)    
            return max(left_depth,right_depth)+1
        if root:
            TreeDepth(root)
        return self.max_sum   

c++程式碼：

/**
 * 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 max_sum=0;
    int diameterOfBinaryTree(TreeNode* root) {
        if(root)
            TreeDepth(root);
        return max_sum;
    }
    int TreeDepth(TreeNode* root)
    {
        int left_depth=0,right_depth=0;
        if (root->left)
            left_depth=TreeDepth(root->left);
        if(root->right)
            right_depth=TreeDepth(root->right);
        max_sum=max(max_sum,left_depth+right_depth);
        return max(left_depth,right_depth)+1;
    }      
};

總結： 養成dfs之前先進行判斷的習慣，這樣可以節省大量的時間～