分析路径走向的三种可能即可
/**
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* };
*/
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param root TreeNode类
* @return int整型
*/
int result = 0;
int dfs(TreeNode* root){
if(root == NULL)
return 0;
int left = dfs(root->left);
int right = dfs(root->right);
result = max(result,max(left,right) + 1);
result = max(result,left + right + 1);
return (left > right ? left : right) + 1;
}
int diameterOfBinaryTree(TreeNode* root) {
// write code here
if(root == NULL)
return 0;
dfs(root);
return result - 1;
}
};