Formatted question description: https://leetcode.ca/all/563.html
563. Binary Tree Tilt (Easy)
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:
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 = 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, 104]. -1000 <= Node.val <= 1000
Related Topics:
Tree, Depth-first Search, Recursion
Similar Questions:
Solution 1.
// OJ: https://leetcode.com/problems/binary-tree-tilt/
// Time: O(N)
// Space: O(H)
class Solution {
int ans = 0;
int dfs(TreeNode *root) {
if (!root) return 0;
int L = dfs(root->left), R = dfs(root->right);
ans += abs(L - R);
return L + R + root->val;
}
public:
int findTilt(TreeNode* root) {
dfs(root);
return ans;
}
};
Java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int findTilt(TreeNode root) {
if (root == null)
return 0;
int leftTilt = findTilt(root.left), rightTilt = findTilt(root.right);
int leftSum = nodeValuesSum(root.left), rightSum = nodeValuesSum(root.right);
int rootTilt = Math.abs(leftSum - rightSum);
return leftTilt + rightTilt + rootTilt;
}
public int nodeValuesSum(TreeNode root) {
if (root == null)
return 0;
else
return root.val + nodeValuesSum(root.left) + nodeValuesSum(root.right);
}
}