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 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.

題意:

找出一個樹的最大直徑，直徑的例子如下:

          1         / \        2   3       / \           4   5    

最大直徑的路徑可為:4->2->1->3或5->2->1->3，這兩個直徑都是3，故返回3

題解:

需要維護一個全局最大直徑，然後再從根結點往下做下去，如下面步驟:

1. 取得左子樹的最大深度與右子樹的最大深度(遞歸的做)
2. 將左右子樹的最大子樹相加作為當前最大直徑，然後維護全局最大直徑
3. 得到當前最大深度(左子樹的最大深度或右子樹的最大深度的比較哪個大)
4. 返回當前最大深度

最後再返回一個全局最大深度

package LeetCode.Easy;import LeetCode.Dependencies.TreeNode;/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class DiameterOfBinaryTree {    public int diameterOfBinaryTree(TreeNode root) {        helper(root);        return global_max_path;    }        int global_max_path = 0;    int helper(TreeNode root) {        if(root == null)            return 0;                //取得左右子樹的最大深度        int left = helper(root.left);        int right = helper(root.right);                //left + right 為左右路徑相加(最長路徑)        global_max_path = Math.max(global_max_path, left + right);                //取得當前最大深度        int cur_max_depth = Math.max(left, right) + 1;                return cur_max_depth;    }}