Leetcode dfs Path SumII

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

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

/*
题意：给定一棵二叉树和一个值，在二叉树中找到从根到叶子的路径使得路径中的节点的总值
等于给定值
思路：dfs
复杂度：时间O(n) 空间O(log n)
*/

/**
 * 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<int> cur;
    	_pathSum(root, sum,cur);
    	return res;
    }
    
private:
    vector<vector<int> > res;
    
    void _pathSum(TreeNode *root, int sum, vector<int> &path){
    	if(!root) return ;
    	path.push_back(root->val);
    	if(!root->left && !root->right){
    		if(root->val == sum) {
    			res.push_back(path);
    		}
    	}
    	_pathSum(root->left, sum - root->val, path);
    	_pathSum(root->right, sum - root->val, path);
    	path.pop_back();
    
    }
};

Leetcode dfs Path SumII

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原文地址：http://blog.csdn.net/zhengsenlie/article/details/38962755