## The challenge#

Given the `root` of a binary tree, return the sum of every tree node’s tilt.

The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values. If a node does not have a left child, then the sum of the left subtree node values is treated as ``. The rule is similar if there the node does not have a right child.

Example 1:

```Input: root = [1,2,3]
Output: 1
Explanation:
Tilt of node 2 : |0-0| = 0 (no children)
Tilt of node 3 : |0-0| = 0 (no children)
Tile 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 = ```

Example 2:

```Input: root = [4,2,9,3,5,null,7]
Output: 15
Explanation:
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```

Example 3:

```Input: root = [21,7,14,1,1,2,2,3,3]
Output: 9```

Constraints:

• The number of nodes in the tree is in the range `[0, 10<sup>4</sup>]`.
• `-1000 <= Node.val <= 1000`

## Definition for a Binary Tree Node#

``````public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
``````

## The solution in Java code#

``````class Solution {
public int findTilt(TreeNode root) {
if (root == null) return 0;

int[] result = walk(root);
return result[1];
}

private int[] walk(TreeNode root) {
if (root == null) return new int[] {0, 0};

int[] left = walk(root.left);
int[] right = walk(root.right);

int subtreeSum = left[0] + right[0] + root.val;
int tiltSum = left[1] + right[1] + Math.abs(left[0] - right[0]);

return new int[] {subtreeSum, tiltSum};
}
}
``````