[LeetCode] Binary Tree Tilt

Problem: 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 0. The rule is similar if there the node does not have a right child.

Example:

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)
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

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

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, 104].
• -1000 <= Node.val <= 1000

Approach: We can use post-order traversal here. If you see it is kind of bottom up as first we need to calculate left and right subtree sum then only we can calculate the tilt value of a node. 

Please directly look at implementation to understand the approach in details as it is fairly easy to understand.

Implementation in C#:

    public int FindTilt(TreeNode root) 

    {

        if (root == null)

        {

            return 0;

        }

        int tiltVal = 0;

        this.FindSumAndGetTilt(root, ref tiltVal);

        return tiltVal;

    }

    

    private int FindSumAndGetTilt(TreeNode node, ref int tilt)

    {

        if (node == null)

        {

            return 0;

        }

        int leftSum = this.FindSumAndGetTilt(node.left, ref tilt);

        int rightSum = this.FindSumAndGetTilt(node.right, ref tilt);

        tilt += Math.Abs(leftSum - rightSum);

        return leftSum + rightSum + node.val;

    }

Complexity: O(n)