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 longestpath 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:题目翻译一下:一个二叉树,要求找二叉树中最远的两个节点之间的距离,距离定义为路径上连边的个数。
两个点的距离,很容易想到最远的两个点肯定在某个树的左右子树上,因此想到计算树的高度。用res来保存以某个节点root为根节点的子树上最远的距离。DFS方法其实也就是对求树的高度算法的改变,还是很容易想到的,代码如下:
1 class Solution { 2 int res = 0; 3 public int diameterOfBinaryTree(TreeNode root) { 4 if ( root == null ) return 0; 5 helper(root); 6 return res-1; 7 } 8 private int helper(TreeNode root) { 9 if ( root == null ) return 0; 10 int left_high = helper(root.left); 11 int right_high = helper(root.right); 12 res = Math.max(res,left_high+right_high+1); 13 // System.out.println(root.val+" "+res); 14 return Math.max(left_high,right_high)+1; 15 } 16 }
我们以上面的例子来讲解一下代码的流程:
4 1 5 1 2 3 3 3 1 4
13行的输出语句格式:第一列是遍历的节点,第二列是res值。可以看到从叶节点向上遍历,先左子树,后右子树。