563. Binary Tree Tilt

1. Question

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.

2. Examples

Example 1:

img
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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:

img
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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:

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Input: root = [21,7,14,1,1,2,2,3,3]
Output: 9

3. Constraints

  • The number of nodes in the tree is in the range [0, 104].
  • -1000 <= Node.val <= 1000

4. References

来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/binary-tree-tilt 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

5. Solutions

这是一道少见的递归结果与最终答案不一样的题目,值得收藏。

QQ20211119-133623
Image
/**
 * 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 {

  int sum = 0;

  public int findTilt(TreeNode root) {
    dfs(root);
    return sum;
  }

  private int dfs(TreeNode root) {
    if (root == null) {
      return 0;
    }

    // 计算当前节点的坡度
    int left = dfs(root.left);
    int right = dfs(root.right);

    // 累加,计算整个树的坡度
    sum += Math.abs(left - right);

    // 返回当前树的所有值供父节点计算坡度
    return root.val + left + right;
  }
}
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