[LeetCode][563] Binary Tree Tilt题解

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[LeetCode][563] Binary Tree Tilt题解

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给定一个树，定义一个树的左右子树的差值的绝对值为树的坡度。
要求返回树的所有节点的坡度的和

dfs搜索左右子树的和 然后相减取绝对值，将这个结果累加到全局变量res上
dfs作用即是记录当前树的总和，以便它的父亲调用（回溯时候调用）

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/*
 * [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;
    }
    //返回的当前节点的左右子树和自己节点的和
    //因此
    //每一次dfs 用temp记录当前的累加值
    //而使用dfs的返回值（当前树的总共的值）作为返回给父亲节点来调用
    int dfs(TreeNode *root)
    {
        if (root == nullptr)
        {
            return 0;
        }
        //如果是叶子节点
        int l=dfs(root->left);
        int r=dfs(root->right);
        res+=abs(l-r);

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

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