LeetCode-Diameter of Binary Tree

今天看到了一个LeetCode上的题目，是一个周赛的第一题，具体的地址为：https://leetcode.com/contest/leetcode-weekly-contest-24/problems/diameter-of-binary-tree/，然后其中的第一题是：

543. Diameter of Binary Tree My SubmissionsBack To Contest

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.

题目大致意思是求一棵树的直径，而这个直径是这棵树中拥有最长路径的两个节点之间的距离，且直径所在的路径并不一定会经过根节点。

上面给出了一个示例，可以看到最长的路径应该是【4，2，1，3】或者【5，2，1，3】，四个点之间的距离只有3，所以返回3。

这题给我最开始的想法其实很简单，就是分别求每个节点的左右两边子树的最长路径，这样就可以得到经过node节点的直径长度

d(node) = max(left) + max(right)

然后将当前的直径与全局保存的直接进行对比，如果全局直径小于这个直径，就更新全局直径的值为经过这个node节点的值。

思路出来了，代码就好写了：

/**
 * 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 diameter = 0;

    int maxLen(TreeNode* root) {
        if (root == NULL) return 0;

        int left = root->left == NULL ? 0 : 1 + maxLen(root->left);
        int right = root->right == NULL ? 0 : 1 + maxLen(root->right);

        if (diameter < left + right) diameter = left + right;

        return left > right ? left : right;

    }

    int diameterOfBinaryTree(TreeNode* root) {

        maxLen(root);

        return diameter;
    }

};