Binary Tree Tilt Problem & Solution
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.
See the binary tree tilt problem on LeetCode.
C++ Solution
#pragma GCC optimize("Ofast")
#pragma GCC optimization("max-inline-insns-recursive-auto")
static const int _=[](){ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();
/**
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int findTilt(TreeNode* root) {
int result = 0;
findTilt(root, result);
return result;
}
private:
int findTilt(TreeNode* root, int& sum) {
if (root == nullptr) {
return 0;
}
if (root->left == nullptr && root->right == nullptr) {
return root->val;
}
int left = findTilt(root->left, sum);
int right = findTilt(root->right, sum);
sum += abs(left - right);
return left + right + root->val;
}
};