Leetcode 113 Path Sum II

triplebee

发布于 2018-01-12 14:53:25

4630

发布于 2018-01-12 14:53:25

文章被收录于专栏：计算机视觉与深度学习基础计算机视觉与深度学习基础

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]
]

找出根到叶和为sum的路径，

在112题基础上多加一个记录路径和结果的变量就行了。

这一段树形的题刷得好无聊。

/**
 * 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 {
public:
    void dfs(TreeNode* root,int now,int sum,vector<vector<int>>& res,vector<int> path)
    {
        now+=root->val;
        path.push_back(root->val);
        if(!root->left && !root->right)
        {
            if(now==sum) res.push_back(path);
            return;
        }
        if(root->left) dfs(root->left,now,sum,res,path);
        if(root->right) dfs(root->right,now,sum,res,path);
    }
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> res;
        vector<int> path;
        if(!root) return res;
        dfs(root,0,sum,res,path);
        return res;
    }
};

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原始发表：2016-10-12 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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