563. Binary Tree Tilt

Description

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 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

Solutions

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int ans;

public int findTilt(TreeNode root) {
ans = 0;
sum(root);
return ans;
}

private int sum(TreeNode root) {
if (root == null) {
return 0;
}
int left = sum(root.left);
int right = sum(root.right);
ans += Math.abs(left - right);
return root.val + left + right;
}
}

• /**
* Definition for a binary tree node.
* 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 ans;

int findTilt(TreeNode* root) {
ans = 0;
sum(root);
return ans;
}

int sum(TreeNode* root) {
if (!root) return 0;
int left = sum(root->left), right = sum(root->right);
ans += abs(left - right);
return root->val + left + right;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def findTilt(self, root: TreeNode) -> int:
ans = 0

def sum(root):
if root is None:
return 0
nonlocal ans
left = sum(root.left)
right = sum(root.right)
ans += abs(left - right)
return root.val + left + right

sum(root)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
var ans int

func findTilt(root *TreeNode) int {
ans = 0
sum(root)
return ans
}

func sum(root *TreeNode) int {
if root == nil {
return 0
}
left, right := sum(root.Left), sum(root.Right)
ans += abs(left - right)
return root.Val + left + right
}

func abs(x int) int {
if x > 0 {
return x
}
return -x
}