<p>Given the <code>root</code> of a binary tree, return <em>the sum of every tree node&#39;s <strong>tilt</strong>.</em></p>

<p>The <strong>tilt</strong> of a tree node is the <strong>absolute difference</strong> between the sum of all left subtree node <strong>values</strong> and all right subtree node <strong>values</strong>. If a node does not have a left child, then the sum of the left subtree node <strong>values</strong> is treated as <code>0</code>. The rule is similar if the node does not have a right child.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/20/tilt1.jpg" style="width: 712px; height: 182px;" />
<pre>
<strong>Input:</strong> root = [1,2,3]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 
Tilt of node 2 : |0-0| = 0 (no children)
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3)
Sum of every tilt : 0 + 0 + 1 = 1
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/20/tilt2.jpg" style="width: 800px; height: 203px;" />
<pre>
<strong>Input:</strong> root = [4,2,9,3,5,null,7]
<strong>Output:</strong> 15
<strong>Explanation:</strong> 
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 5 : |0-0| = 0 (no children)
Tilt of node 7 : |0-0| = 0 (no children)
Tilt of node 2 : |3-5| = 2 (left subtree is just left child, so sum is 3; right subtree is just right child, so sum is 5)
Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7)
Tilt of node 4 : |(3+5+2)-(9+7)| = |10-16| = 6 (left subtree values are 3, 5, and 2, which sums to 10; right subtree values are 9 and 7, which sums to 16)
Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/20/tilt3.jpg" style="width: 800px; height: 293px;" />
<pre>
<strong>Input:</strong> root = [21,7,14,1,1,2,2,3,3]
<strong>Output:</strong> 9
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    ans=0
    def findTilt(self, root: Optional[TreeNode]) -> int:
        self.ans=0
        def f(root):
            if root:
                le=f(root.left)
                ri=f(root.right)
                ti=abs(le-ri)
                self.ans+=ti
                # print(root.val,le,ri,ti,self.ans)
                return root.val+le+ri
            else:
                return 0
        f(root)
        return self.ans
17500