LeetCode.543 Diameter of Binary Tree

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 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.

分析（原创）：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int max=0;
    public int diameterOfBinaryTree(TreeNode root) {
        //给定二叉树，求任意两个节点之间最长的距离
        //思路：分别递归求出左右子树长度，最后为俩长度之和（可能是其中一段，不经过根节点，所以设置一个全局变量，记录最大值）。
        
        //出口
        if(root==null) return 0;
        //过程
        int result=treeLength(root.left)+treeLength(root.right);
        //递归
        int left=diameterOfBinaryTree(root.left);
        int right=diameterOfBinaryTree(root.right);
        
        //最后结果
        max=Math.max(max,Math.max(result,Math.max(left,right)));
        return max;
    }
    public int treeLength(TreeNode root){
        if(root==null) return 0;
        int left=treeLength(root.left)+1;
        int right=treeLength(root.right)+1;
        return Math.max(left,right);
    }
}

分析2（参考答案）：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int max=0;
    public int diameterOfBinaryTree(TreeNode root) {
         //给定二叉树，求任意两个节点之间最长的距离
        //思路：分别递归求出左右子树长度，最后为俩长度之和（可能是其中一段，不经过根节点，所以设置一个全局变量，记录最大值）。
        treeLength(root);
        return max;
    }
    //递归
    public int treeLength(TreeNode root){
        if(root==null)  return 0;
        //递归求解树左右长度
        int left=treeLength(root.left);
        int right=treeLength(root.right);
        
        max=Math.max(max,left+right);
        return Math.max(left,right)+1;
        
    }
}