[leetcode] 543. Diameter of Binary Tree

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 thelongest 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.

这道题是计算二叉树“直径”，题目难度为Easy。

这道题和第124题非常相近，感兴趣的同学可以顺便看下第124题（传送门）。

所谓的“直径”，即任意两节点之间的最大距离。“直径”所在的路径必定存在一个自己的根节点将其分为左右两部分，两部分分别是其子节点最大深度对应的节点路径，这样题目就转换为计算节点深度了。知道了两个子节点的深度之后就可以计算出以此节点为根的“直径”，通过深度优先遍历二叉树即可得出最终最大的“直径”。具体代码：

class Solution {
    int getRadius(TreeNode* n, int& diameter) {
        int lRadius = 0, rRadius = 0;
        if(n->left) lRadius = getRadius(n->left, diameter);
        if(n->right) rRadius = getRadius(n->right, diameter);
        diameter = max(diameter, lRadius+rRadius);
        return max(lRadius, rRadius) + 1;
    }
public:
    int diameterOfBinaryTree(TreeNode* root) {
        if(!root) return 0;
        int diameter = INT_MIN;
        getRadius(root, diameter);
        return diameter;
    }
};