C++简单的递归+回溯

二叉树中和为某一值的路径(二)

http://www.nowcoder.com/practice/b736e784e3e34731af99065031301bca

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    vector<vector<int>> ans;
    vector<int> v;
    void dfs(TreeNode *root, int sum) {
        if(!root) return;
        v.push_back(root->val);
        if(!root->left && !root->right && sum == root->val)
            ans.push_back(v);
        dfs(root->left, sum - root->val);
        dfs(root->right, sum - root->val);
        v.pop_back();
    }
    vector<vector<int>> FindPath(TreeNode* root,int expectNumber) {
        dfs(root, expectNumber);
        return ans;
    }
};
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