题目： 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.
解答： 注意点： 1、树的直径中的路径不一定过根节点 2、路径的长度是代表路径中边的数目。 那么可以想一下二叉树深度的递归代码，如何将它应用到我们这个题目中来：
//求二叉树深度的代码 int dfs(TreeNode *root){ if(root==NULL) return 0; int left=dfs(root->left); int right=dfs(root->right); return max(left,right)+1; }不过以上代码是以根节点为起点出发的，而根据题意，二叉树的直径路径可能不经过根节点，那么对于每个节点，我们遍历它左右子树的路径长度之后，将其累加，与一个当前的最大直径长度比较。代码如下：
/** * 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 diameterOfBinaryTree(TreeNode* root) { if(root==NULL) return 0; int maxD=0; maxHeight(root,maxD); return maxD; } int maxHeight(TreeNode *root,int &maxD){ if(root==NULL) return 0; int left=maxHeight(root->left,maxD); int right=maxHeight(root->right,maxD); maxD=max(maxD,left+right); return max(left,right)+1; //这里注意是路径长度 } };