【Leetcode】Path Sum II

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：与【Leetcode】Path Sum 不同的是。此题须要保存全部的路径，因此，搜索完左子树的路径后应该继续搜索右子树的路径。

/**
 * 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) {
        vector<vector<int>> result;
        vector<int> path;
        pathSumHelper(result, path, root, sum);
        
        return result;
    }
    
private:
    void pathSumHelper(vector<vector<int>> &result, vector<int> &path, TreeNode *node, int sum)
    {
        if(node == NULL)    return;
        if(node->left == NULL && node->right == NULL && node->val == sum)
        {
            path.push_back(node->val);
            result.push_back(path);
            path.pop_back();
            return;
        }
        
        path.push_back(node->val);
        pathSumHelper(result, path, node->left, sum - node->val);
        pathSumHelper(result, path, node->right, sum - node->val);
        path.pop_back();
    }
};

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