563. Binary Tree Tilt(C语言)

Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes' tilt.

Example:

Input:          1       /   \      2     3Output: 1Explanation: Tilt of node 2 : 0Tilt of node 3 : 0Tilt of node 1 : |2-3| = 1Tilt of binary tree : 0 + 0 + 1 = 1

Note:

1. The sum of node values in any subtree won't exceed the range of 32-bit integer.
2. All the tilt values won't exceed the range of 32-bit integer.

我理解的题意是：

1的下面是2和3,1的tilt是|2-3|=1

2的下面是4和5,2的tilt是|4-5|=1

3的下面是6和7,3的tilt是|6-7|=1

4,5,6,7下面都是NULL，4,5,6,7的tilt是0

所以这个树的tilt是1+1+1+0*4=3

如果你也是这么理解的，那么你和我一样，都理解错了。

题意是：

4,5,6,7下面都是NULL，4,5,6,7的tilt是0

2的下面是4和5,2的tilt是|4-5|=1

3的下面是6和7,3的tilt是|6-7|=1

1的下面是2+4+5和3+6+7,1的tilt是|2+4+5-(3+6+7)|=5

这个树的tilt就是1+1+5=7

我才发现这有个，自己输入测试用例

这样会出来答案7，验证一下

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     struct TreeNode *left; *     struct TreeNode *right; * }; */int findTilt(struct TreeNode* root) {    if(root==NULL){        return 0;    }    return abs(sum(root->left)-sum(root->right))+findTilt(root->left)+findTilt(root->right);}int sum(struct TreeNode* root){    if(root==NULL){        return 0;    }    return sum(root->left)+sum(root->right)+root->val;}