[LeetCode] 563. Binary Tree Tilt

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)
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

二叉树的坡度。

给定一个二叉树，计算 整个树 的坡度 。

一个树的 节点的坡度 定义即为，该节点左子树的节点之和和右子树节点之和的 差的绝对值 。如果没有左子树的话，左子树的节点之和为 0 ；没有右子树的话也是一样。空结点的坡度是 0 。

整个树 的坡度就是其所有节点的坡度之和。

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。

思路是后序遍历。这道题对于坡度的定义说的很清楚了，但是不知道为什么downvote很多。这里我们需要一个全局变量来统计所有的坡度的和，对于当前节点来说，坡度 = 左子树的坡度 + 右子树的坡度。往父节点返回的是自己的坡度 + 左子树的坡度 + 右子树的坡度。

时间O(n)

空间O(n)

Java实现

/**
 * 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 res = 0;

    public int findTilt(TreeNode root) {
        helper(root);
        return res;
    }

    private int helper(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = helper(root.left);
        int right = helper(root.right);
        res += Math.abs(left - right);
        return left + right + root.val;
    }
}

LeetCode 题目总结