题目

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:

image.png

  1. Input: root = [1,2,3]
  2. Output: 1
  3. Explanation: 
  4. Tilt of node 2 : |0-0| = 0 (no children)
  5. Tilt of node 3 : |0-0| = 0 (no children)
  6. 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)
  7. Sum of every tilt : 0 + 0 + 1 = 1

Example 2:

image.png

  1. Input: root = [4,2,9,3,5,null,7]
  2. Output: 15
  3. Explanation: 
  4. Tilt of node 3 : |0-0| = 0 (no children)
  5. Tilt of node 5 : |0-0| = 0 (no children)
  6. Tilt of node 7 : |0-0| = 0 (no children)
  7. 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)
  8. Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7)
  9. 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)
  10. Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15

Example 3:

image.png

  1. Input: root = [21,7,14,1,1,2,2,3,3]
  2. Output: 9

Constraints:

  • The number of nodes in the tree is in the range [0, 10^4].
  • -1000 <= Node.val <= 1000

题意

统计一个数中所有子树的 |左子树的和-右子树的和| 的和。

思路

直接递归计算每个子树的和再进行统计即可。

代码实现

Java

  1. class Solution {
  2.     private int sum;
  4.     public int findTilt(TreeNode root) {
  5.         sum = 0;
  6.         findSum(root);
  7.         return sum;
  8.     }
  10.     public int findSum(TreeNode root) {
  11.         if (root == null) {
  12.             return 0;
  13.         }
  15.         int sumLeft = findSum(root.left);
  16.         int sumRight = findSum(root.right);
  17.         sum += Math.abs(sumLeft - sumRight);
  18.         return sumLeft + sumRight + root.val;
  19.     }
  20. }