Binary Tree Tilt

LeetCode 563 | Difficulty: Easy

Easy

Problem 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, 10^4]`.

- `-1000 <= Node.val <= 1000`

Topics: Tree, Depth-First Search, Binary Tree

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Approach

Tree DFS

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left → go right → combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

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Solutions

Solution 1: C# (Best: 157 ms)

Metric	Value
Runtime	157 ms
Memory	28.7 MB
Date	2021-10-01

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int result = 0;
    public int FindTilt(TreeNode root) {
    NodeSum(root);
    return result;
}

public int NodeSum(TreeNode root)
{
    if(root == null)
    {
        return 0;
    }
    int leftSum = root.left != null ? NodeSum(root.left) : 0;
    int rightSum = root.right != null ? NodeSum(root.right) : 0;
    result += Math.Abs(leftSum - rightSum);
    return leftSum + rightSum + root.val;
}
}

📜 2 more C# submission(s)

Submission (2021-10-01) — 168 ms, 28.9 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int FindTilt(TreeNode root) {
    int result = 0;
    NodeSum(root, ref result);
    return result;
}

public int NodeSum(TreeNode root, ref int result)
{
    if(root == null)
    {
        return 0;
    }
    int leftSum = root.left != null ? NodeSum(root.left, ref result) : 0;
    int rightSum = root.right != null ? NodeSum(root.right, ref result) : 0;
    result += Math.Abs(leftSum - rightSum);
    return leftSum + rightSum + root.val;
}
}

Submission (2021-10-01) — 186 ms, 28.8 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int FindTilt(TreeNode root) {
       List<int> result = new List<int>();
    NodeSum(root, result);
    return result.Sum();
}

public int NodeSum(TreeNode root, List<int> result)
{
    if(root == null)
    {
        return 0;
    }
    int leftSum = root.left != null ? NodeSum(root.left, result) : 0;
    int rightSum = root.right != null ? NodeSum(root.right, result) : 0;
    result.Add(Math.Abs(leftSum-rightSum));
    return leftSum + rightSum + root.val;
}
}

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Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$

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Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.
• LeetCode provides 4 hint(s) for this problem — try solving without them first.

💡 Hints

Hint 1: Don't think too much, this is an easy problem. Take some small tree as an example.

Hint 2: Can a parent node use the values of its child nodes? How will you implement it?

Hint 3: May be recursion and tree traversal can help you in implementing.

Hint 4: What about postorder traversal, using values of left and right childs?