LeetCode 543. Diameter of Binary Tree 二叉树的直径 (C++/Java)

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.

分析：

给定一棵二叉树，计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过根结点。

我们定义空结点的直径长度为0，而除root结点外，其余所有结点都有一条边连接其父结点。那么递归求解此问题，当前结点所有的直径长度等于其左右孩子的长度之和，也就是把当前结点当成桥接两个孩子结点的桥梁，更新全局的最大值，但如果当前结点还有父结点的话，则应该是左右孩子的长度取最大值加1,1也就是当前结点连向父结点的那条边，而由于题目规定，我们只能通过一次结点，所以取最大值就好。

程序：

C++

/**
 * 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) {
        int res = 0;
        dTree(root, res);
        return res;
    }
private:
    int dTree(TreeNode* root, int& res){
        if(root == nullptr)
            return 0;
        int l = dTree(root->left, res);
        int r = dTree(root->right, res);
        res = max(res, l+r);
        return max(l, r) + 1;
    }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int diameterOfBinaryTree(TreeNode root) {
        res = 0;
        dTree(root);
        return res;
    }
    private int dTree(TreeNode root){
        if(root == null)
            return 0;
        int l = dTree(root.left);
        int r = dTree(root.right);
        res = Math.max(res, l+r);
        return Math.max(l, r) + 1;
    }
    private int res;
}