LeetCode - Easy - 563. Binary Tree Tilt-蒲公英云

LeetCode - Easy - 563. Binary Tree Tilt

红太狼 2022-11-07 15:55 128阅读 0赞

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

运用二叉树的后序遍历算法。

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)));
    }
}