[LeetCode] 563. Binary Tree Tilt
作者：互联网
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
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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
1 /**
2 * Definition for a binary tree node.
3 * public class TreeNode {
4 * int val;
5 * TreeNode left;
6 * TreeNode right;
7 * TreeNode() {}
8 * TreeNode(int val) { this.val = val; }
9 * TreeNode(int val, TreeNode left, TreeNode right) {
10 * this.val = val;
11 * this.left = left;
12 * this.right = right;
13 * }
14 * }
15 */
16 class Solution {
17 int res = 0;
18
19 public int findTilt(TreeNode root) {
20 helper(root);
21 return res;
22 }
23
24 private int helper(TreeNode root) {
25 if (root == null) {
26 return 0;
27 }
28 int left = helper(root.left);
29 int right = helper(root.right);
30 res += Math.abs(left - right);
31 return left + right + root.val;
32 }
33 }
标签：node,Binary,right,TreeNode,Tilt,563,Tree,root,left 来源： https://www.cnblogs.com/cnoodle/p/14615557.html