lintcode596:Minimum Subtree

问题描述

https://www.lintcode.com/problem/minimum-subtree/description

java实现

/**
 * 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 {
    /**
     * @param root: the root of binary tree
     * @return: the root of the minimum subtree
     */
    /*
    分治+全局变量监控的遍历
    用分治每次的确计算出当前的sum并作为上层需要的数据返回。
    但每次helper里同时暗搓搓比一下自己是不是比全局变量优秀了，取而代之。
     */
    private TreeNode subtree=null;
    private int subtreeSum=Integer.MAX_VALUE;
    public TreeNode findSubtree(TreeNode root) {
        helper(root);
        return subtree;
    }
    //定义 ：返回以root为根的和
    private int helper(TreeNode root){
        if(root==null)
            return 0;
        //分治
        int sum=helper(root.left)+helper(root.right)+root.val;
        //遍历
        if(sum<subtreeSum){
            subtreeSum=sum;
            subtree=root;
        }
        return sum;
    }
}