## Description

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.

## Solutions

主要是题意的理解，倾斜度的计算是要左子树和右子树所有节点值的和的差值。

### 1. Recurrence

# Time: O(nlogn)
# Space: O(n)
# 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: TreeNode) -> int:
if not root:
return 0
return abs(self.nodeSum(root.left) - self.nodeSum(root.right)) + self.findTilt(root.left) + self.findTilt(root.right)

def nodeSum(self, root):
if not root:
return 0
return self.nodeSum(root.left) + self.nodeSum(root.right) + root.val

# 75/75 cases passed (596 ms)
# Your runtime beats 7.44 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (14.4 MB)


看了 submission，最快的把这两个过程放在一起了，真是厉害！

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

class Solution(object):
def findTilt(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root: return 0
return self.helper(root)[1]

def helper(self, root):
ltot = rtot = ltlt = rtlt = 0
if root.left:
ltot, ltlt = self.helper(root.left)
if root.right:
rtot, rtlt = self.helper(root.right)
return (ltot + rtot + root.val, abs(ltot - rtot) + ltlt + rtlt)