LeetCode OJ：Path Sum II（路径和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]
]

求路径的和与某一特定的值时候对等即可，简单的dfs，代码如下：

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<int> res;
        dfs(root, res, sum);
        return ret;
    }

    void dfs(TreeNode * root, vector<int> res, int left)
    {
        if(!root) return;
        if(!root->left && !root->right && left == root->val){
            res.push_back(root->val);
            ret.push_back(res);
        }
        if(left <= root->val)
            return;
        else{
            res.push_back(root->val);
            if(root->left)
                dfs(root->left, res, left -= root->val);
            if(root->right)
                dfs(root->right, res, left -= root->val);
        }
    }
private:
    vector<vector<int>> ret;
};

 java版本代码如下，方法相同，就是java的引用处理起来稍微麻烦点，递归尾部应该pop一下。

public class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> ret = new ArrayList<List<Integer>>();
        Stack<Integer> tmp = new Stack<Integer>();
           dfs(ret, tmp, root, sum);
           return ret;
    }

    public void dfs(List<List<Integer>> ret, Stack<Integer> tmp, TreeNode root, int remain){
        if(root == null) return;
        tmp.push(root.val);
        if(root.left == null && root.right == null)
            if(remain == root.val)
                ret.add((Stack<Integer>)tmp.clone());
        if(root.left != null) dfs(ret, tmp, root.left, remain - root.val);
        if(root.right != null) dfs(ret, tmp, root.right, remain - root.val);
        tmp.pop();
    }
}