563 Binary Tree Tilt

1. 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 sum of all left subtree 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 doe not have a right child.

 

constraints :

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

 

2. Algorithms

Let's see when we got [4, 2, 9, 3, 5, null, 7] as input. Then our result of algorithm will be : | 0 - 0 | + | 0 - 0 | + | 0 - 0 | + | 3 - 5 | + | 0 - 7 | + | ( 3 + 5 + 2 ) - ( 9 + 7 ) | = 0 + 0 + 0 + 2 + 7 + 6 = 15.

 

1. Start
2. Declare variable named sum_.
3. Initilalize sum_ to 0.
4. Declare function named helper which get root as input.
5. Set num_ as nonlocal.
6. (Base case) If root is None, return 0.
7. Traverse left subtree recursively.
8. Traverse right subtree recursively.
9. Add up absolute difference between left value and right value.
10. Return sum between left and right and value which sum of all left subtree values and all right subtree node values recursively.
11. End

 

3. Codes

# 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: Optional[TreeNode]) -> int:
        sum_ = 0 
        def helper(root) : 
            nonlocal sum_ 
            if root is None : 
                return 0 
            left = helper(root.left)
            right = helper(root.right)
            sum_ += abs(left - right) 
            
            return left + right + root.val 
        helper(root)
        return sum_

 

4. Conclusion