Question

563. Binary Tree Tilt

中文文档

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

Solution 1

# 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.
 * 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.
 * 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
}