LeetCode 543. Diameter of Binary Tree

Description
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.

Analysis
这道题我做了很久，原因是因为我一直想错了方向。
题意是要就我们找到最长路，一开始我想着如果要求最长路那么一定是经过根，然后求最长的深度左右子树的最长深度加起来就可以了。
后来才发现我没读懂题目，我一直以为题目中的不经过根没什么特别的意思，其实是有意思的。
我最后的做法就是利用递归，然后每次都将diameter 和 left+right来更新diameter最后得到结果。

Code

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; *///wrong class Solution {public:    int depthTree(TreeNode* root){        if(root == NULL) return 0;        //if(root->left==NULL&& root->right==NULL) return 1;        TreeNode* right = root->right;        TreeNode* left = root->left;        return 1+max(depthTree(right),depthTree(left));    }    int diameterOfBinaryTree(TreeNode* root) {        if(root == NULL) return 0;        if(root->left==NULL&& root->right==NULL) return 0;        TreeNode* right = root->right;        TreeNode* left = root->left;       // if(root->left!=NULL&& root->right!=NULL) return 2+depthTree(left)+depthTree(right);        /*else*/ return depthTree(right)+depthTree(left);    }};//correctclass Solution {public:    int depthTree(TreeNode* root, int& diameter) {        if(root == NULL) return 0;        int left=depthTree(root->left, diameter);        int right=depthTree(root->right, diameter);        diameter = max(diameter, left+right);        return 1+max(left,right);    }    int diameterOfBinaryTree(TreeNode* root) {        if(root == NULL) return 0;        int diameter = 0;        depthTree(root, diameter);        return diameter;    }};