Formatted question description: https://leetcode.ca/all/563.html

563. Binary Tree Tilt (Easy)

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)
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)
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

Related Topics:
Tree, Depth-first Search, Recursion

Similar Questions:

Solution 1.

// OJ: https://leetcode.com/problems/binary-tree-tilt/
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    int dfs(TreeNode *root) {
        if (!root) return 0;
        int L = dfs(root->left), R = dfs(root->right);
        ans += abs(L - R);
        return L + R + root->val;
    }
public:
    int findTilt(TreeNode* root) {
        dfs(root);
        return ans;
    }
};

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public int findTilt(TreeNode root) {
            if (root == null)
                return 0;
            int leftTilt = findTilt(root.left), rightTilt = findTilt(root.right);
            int leftSum = nodeValuesSum(root.left), rightSum = nodeValuesSum(root.right);
            int rootTilt = Math.abs(leftSum - rightSum);
            return leftTilt + rightTilt + rootTilt;
        }
    
        public int nodeValuesSum(TreeNode root) {
            if (root == null)
                return 0;
            else
                return root.val + nodeValuesSum(root.left) + nodeValuesSum(root.right);
        }
    }
    
  • // OJ: https://leetcode.com/problems/binary-tree-tilt/
    // Time: O(N)
    // Space: O(H)
    class Solution {
        int ans = 0;
        int dfs(TreeNode *root) {
            if (!root) return 0;
            int L = dfs(root->left), R = dfs(root->right);
            ans += abs(L - R);
            return L + R + root->val;
        }
    public:
        int findTilt(TreeNode* root) {
            dfs(root);
            return ans;
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
      def findTilt(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
    
        def dfs(root):
          if not root:
            return 0, 0  # sum, tilt sum
    
          leftSum, leftTilt = dfs(root.left)
          rightSum, rightTilt = dfs(root.right)
    
          return leftSum + root.val + rightSum, abs(leftSum - rightSum) + leftTilt + rightTilt
    
        return dfs(root)[1]
    
    

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