Description

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

tree

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.

Solution

DFS

以每个节点为起点，往左右两边扩展，更新diameter。

这里用到一个depth方法用作辅助，但是这个depth的定义跟惯用的定义不太一样，对于leaf节点depth是1，对于root节点则是root到最深leaf的距离+1。

这道题唯一要注意的就是值不要弄错了。

/**
 * 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) {
        if (root == null) {
            return 0;
        }
        
        int path = depth(root.left) + depth(root.right);
        int leftPath = diameterOfBinaryTree(root.left);
        int rightPath = diameterOfBinaryTree(root.right);
        
        return Math.max(path, Math.max(leftPath, rightPath));
    }
    
    public int depth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        return 1 + Math.max(depth(root.left), depth(root.right));
    }
}