LeetCode | Path Sum II

2023年5月10日 188次阅读 来源: Allanxl

题目：

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.

For example:

Given the below binary tree and 
sum = 22
,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

思路：

利用递归的方法，当左右子节点都是NULL的时候能够判断当前节点是叶子节点。类似http://blog.csdn.net/lanxu_yy/article/details/11787805

代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> pathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        vector<vector<int>>*  v = new vector<vector<int>>();
        
        if(root == NULL)
        {
            return *v;
        }
        
        vector<int> tmp;
        pathSum(root, 0, sum, tmp, v);
        return *v;
    }
    
    void pathSum(TreeNode * p, int total, int sum, vector<int> tmp, vector<vector<int>> * v)
    {
        if((p->left == NULL) && (p->right == NULL))
        {
            if(total + p->val  == sum)
            {
                tmp.push_back(p->val);
                v->push_back(tmp);
            }
        }
        else if(p->left == NULL)
        {
            tmp.push_back(p->val);
            pathSum(p->right, p->val + total, sum, tmp, v);
        }
        else if(p->right == NULL)
        {
            tmp.push_back(p->val);
            pathSum(p->left, p->val + total, sum, tmp, v);
        }
        else
        {
            tmp.push_back(p->val);
            pathSum(p->left, p->val + total, sum, tmp, v);
            pathSum(p->right, p->val + total, sum, tmp, v);
        }
    }
};
    原文作者：Allanxl
    原文地址: https://blog.csdn.net/lanxu_yy/article/details/11787951
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
点赞