597.具有最大平均数的子树

描述

给一棵二叉树，找到有最大平均值的子树。返回子树的根结点。

注意事项

LintCode会打印出根结点为你返回节点的子树，保证有最大平均数子树只有一棵

样例

给一个二叉树：

     1
   /   \
 -5     11
 / \   /  \
1   2 4    -2 

返回节点 11。

代码

Traverse + Divide Conquer  
 /**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    private class ResultType {
        public int size;
        public int sum;
        public ResultType (int sum, int size) {
            this.sum = sum;
            this.size = size;
        }
    }
    
    // subtree代表的是当前最大平均数的根结点
    // subtreeResult代表当前最大平均数的ResultType
    private TreeNode subtree = null;
    private ResultType subtreeResult = null;
    
    public TreeNode findSubtree2(TreeNode root) {
        helper(root);
        return subtree;
    }
    
    private ResultType helper(TreeNode root) {
        if (root == null) {
            return new ResultType(0, 0);
        }
        
        ResultType left = helper(root.left);
        ResultType right = helper(root.right);
        // result代表当前结点开头的二叉树的数值之和
        ResultType result = new ResultType(
            left.sum + right.sum + root.val,
            left.size + right.size + 1
        );
      
        /* 从subtreeResult.sum / subtreeResult.size < 
         * result.sum / result.size写成下面两端相乘形式来减小精度误差
         */
        // result结果和全局最优比较，全局最优是空时直接换成result
        if (subtree == null || 
            result.sum * subtreeResult.size > subtreeResult.sum * result.size) {
            subtree = root;
            subtreeResult = result;
        }
        return result;
    } 
}