Path Sum II

最新推荐文章于 2024-10-26 02:26:14 发布

weixin_34234721

最新推荐文章于 2024-10-26 02:26:14 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/winscoder/p/3535398.html

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]
]

Code:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void findPath(TreeNode *node, int n, int &sum, vector<int> &buf, vector<vector<int>> &res){
        if(!node)
            return;
        buf.push_back(node->val);
        if(!node->left&&!node->right&&node->val+n==sum){
            res.push_back(buf);
            buf.pop_back();
            return;
        }
        findPath(node->left,node->val+n,sum,buf,res);
        findPath(node->right,node->val+n,sum,buf,res);
        buf.pop_back();
    }
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
            vector<int> buf;
            vector<vector<int>> res;
            findPath(root,0,sum,buf,res);
            return res;
    }
};

转载于:https://www.cnblogs.com/winscoder/p/3535398.html