LeetCode 563. Binary Tree Tilt

最新推荐文章于 2025-08-24 09:29:29 发布

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最新推荐文章于 2025-08-24 09:29:29 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/A-Little-Nut/p/10073981.html

本文介绍了一种通过后序遍历算法计算二叉树每个节点的左子树和右子树值之差的绝对值（即倾斜），并返回整个二叉树的总倾斜值的方法。该方法首先递归地计算左右子树的和，然后计算当前节点的倾斜值，并累加到全局倾斜值中。

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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 {// 后序遍历求和
public:
    int postorder(TreeNode* root){
        if(root==NULL) return 0;
        int leftsum=postorder(root->left);
        int rightsum=postorder(root->right);
        sum+=abs(leftsum-rightsum);
        return leftsum+rightsum+root->val;
    }
    int findTilt(TreeNode* root) {
        postorder(root);
        return sum;
    }
private:
    int sum=0;
};

转载于:https://www.cnblogs.com/A-Little-Nut/p/10073981.html