Topic
- Tree
- Depth-first Search
- Recursion
Description
https://leetcode.com/problems/binary-tree-tilt/
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)
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:
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
Analysis
Use the post-order traversal algorithm of the binary tree.
Submission
import com.lun.util.BinaryTree.TreeNode;
public class BinaryTreeTilt {
public int findTilt(TreeNode root) {
int[] sum = {
0};
findTilt(root, sum);
return sum[0];
}
private int findTilt(TreeNode root, int[] sum) {
if (root == null) return 0;
int leftSum = findTilt(root.left, sum);
int rightSum = findTilt(root.right, sum);
sum[0] += Math.abs(leftSum - rightSum);
return leftSum + rightSum + root.val;
}
}
Test
import static org.junit.Assert.*;
import org.junit.Test;
import com.lun.util.BinaryTree;
public class BinaryTreeTiltTest {
@Test
public void test() {
BinaryTreeTilt obj = new BinaryTreeTilt();
assertEquals(1, obj.findTilt(BinaryTree.integers2BinaryTree(1, 2, 3)));
assertEquals(15, obj.findTilt(BinaryTree.integers2BinaryTree(4, 2, 9, 3, 5, null, 7)));
assertEquals(9, obj.findTilt(BinaryTree.integers2BinaryTree(21, 7, 14, 1, 1, 2, 2, 3, 3)));
}
}