leetcod刷题（19）——563. 二叉树的坡度

输入: 
         1
       /   \
      2     3
输出: 1
解释: 
结点的坡度 2 : 0
结点的坡度 3 : 0
结点的坡度 1 : |2-3| = 1
树的坡度 : 0 + 0 + 1 = 1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int val=0;
    public int findTilt(TreeNode root) {
        find(root);
        return val;
    }
    
    public int find(TreeNode root){
        if(root ==null){
            return 0;
        }
        int left = find(root.left);
        int right = find(root.right);
        val+= Math.abs(left-right);
        return left+right+root.val;
    }
}