二叉树的坡度（左子树节点和-右子树节点和）Binary Tree Tilt

问题：

Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes' tilt.

Example:

Input: 
         1
       /   \
      2     3
Output: 1
Explanation: 
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1

Note:

1. The sum of node values in any subtree won't exceed the range of 32-bit integer.
2. All the tilt values won't exceed the range of 32-bit integer.

解决：

【注】左右子树的坡度 = | 左子树全部节点的值的和 - 右子树全部节点的值的和 |

① 本题要求二叉树的坡度，某个结点的坡度的定义为该结点的左子树之和与右子树之和的差的绝对值，这道题让咱们求全部结点的坡度之和。

采用后序遍历，左子节点---右子节点---根节点的顺序，能够由叶节点开始处理，能够比较方便的计算出每一个节点的累加和，同时也能够根据子树的和来计算坡度。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution { // 10 ms
    int tilt = 0;
    public int findTilt(TreeNode root) { 
        postorder(root);
        return tilt;
    }
    public int postorder(TreeNode node){
        if(node == null) return 0;
        int lsum = postorder(node.left);
        int rsum = postorder(node.right);
        tilt += Math.abs(lsum - rsum);
        return lsum + rsum + node.val; //返回的是当前节点做为根节点的子树的和     } }