[LeetCode][563] Binary Tree Tilt題解

給定一個樹，定義一個樹的左右子樹的差值的絕對值為樹的坡度。
要求返回樹的所有節點的坡度的和

dfs搜尋左右子樹的和 然後相減取絕對值，將這個結果累加到全域性變數res上
dfs作用即是記錄當前樹的總和，以便它的父親呼叫（回溯時候呼叫）

/*
 * [563] Binary Tree Tilt
 *
 * https://leetcode.com/problems/binary-tree-tilt/description/
 *
 * algorithms
 * Easy (46.46%)
 * Total Accepted:    39.4K
 * Total Submissions: 84.9K
 * Testcase Example:  '[1,2,3]'
 *
 * 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     3
 * Output: 1
 * Explanation:
 * Tilt of node 2 : 0
 * Tilt of node 3 : 0
 * Tilt of node 1 : |2-3| = 1
 * Tilt of binary tree : 0 + 0 + 1 = 1
 *
 *
 *
 * Note:
 *
 * The sum of node values in any subtree won't exceed the range of 32-bit
 * integer.
 * All the tilt values won't exceed the range of 32-bit integer.
 *
 *
 */
 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
  private:
    int res;

  public:
    int findTilt(TreeNode *root)
    {
        res = 0
 
;
        dfs(root);
        return res;
    }
    //返回的當前節點的左右子樹和自己節點的和
    //因此
    //每一次dfs 用temp記錄當前的累加值
    //而使用dfs的返回值（當前樹的總共的值）作為返回給父親節點來呼叫
    int dfs(TreeNode *root)
    {
        if (root == nullptr)
        {
            return 0;
        }
        //如果是葉子節點
        int l=dfs(root->left);
        int
 
 r=dfs(root->right);
        res+=abs(l-r);

        return l+r+root->val;
    }
};

參考https://leetcode.com/problems/binary-tree-tilt/discuss/102323/C+±easy-and-clean-soluion