Binary Tree Tilt Problem


Description

LeetCode Problem 563.

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 1:

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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

Example 2:

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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:

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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, 10^4].
  • -1000 <= Node.val <= 1000


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class Solution {
public:
    int tiltSum = 0;
    int findTilt(TreeNode* root) {
        DFS(root);
        return tiltSum;
    }
    int DFS(TreeNode* root) {
        if (!root) return 0;
        int leftSum = DFS(root -> left);                // sum of left subtree
        int rightSum = DFS(root -> right);              // sum of right subtree
        tiltSum += abs(leftSum - rightSum);             // add tilt of current node to overall tiltSum
        return leftSum + rightSum + root -> val;        // returns sum of subtree starting at 'root'
    }
};




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