# 563. Binary Tree Tilt 二叉树的坡度

@TOC

## # 题目描述

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 1:

``````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.

## # 解题方法

### # Java解法

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

### # Python解法

``````# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
def findTilt(self, root):
"""
:type root: TreeNode
:rtype: int
"""
self.sums = 0
self.postOrder(root)
return self.sums

def postOrder(self, root):
if not root:
return 0
left = self.postOrder(root.left)
right = self.postOrder(root.right)
self.sums += abs(left - right)
return left + right + root.val
``````

## # 日期

2017 年 5 月 9 日 2018 年 11 月 16 日 —— 又到周五了！