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[LeetCode] 113. Path Sum II 二叉树路径之和之二

 

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]
]

这道二叉树路径之和在之前那道题 Path Sum 的基础上又需要找出路径,但是基本思想都一样,还是需要用深度优先搜索 DFS,只不过数据结构相对复杂一点,需要用到二维的 vector,而且每当 DFS 搜索到新结点时,都要保存该结点。而且每当找出一条路径之后,都将这个保存为一维 vector 的路径保存到最终结果二维 vector 中。并且,每当 DFS 搜索到子结点,发现不是路径和时,返回上一个结点时,需要把该结点从一维 vector 中移除,参见代码如下:

 

解法一:

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> res;
        vector<int> out;
        helper(root, sum, out, res);
        return res;
    }
    void helper(TreeNode* node, int sum, vector<int>& out, vector<vector<int>>& res) {
        if (!node) return;
        out.push_back(node->val);
        if (sum == node->val && !node->left && !node->right) {
            res.push_back(out);
        }
        helper(node->left, sum - node->val, out, res);
        helper(node->right, sum - node->val, out, res);
        out.pop_back();
    }
};

 

下面这种方法是迭代的写法,用的是中序遍历的顺序,参考之前那道 Binary Tree Inorder Traversal,中序遍历本来是要用栈来辅助运算的,由于要取出路径上的结点值,所以用一个 vector 来代替 stack,首先利用 while 循环找到最左子结点,在找的过程中,把路径中的结点值都加起来,这时候取出 vector 中的尾元素,如果其左右子结点都不存在且当前累加值正好等于 sum 了,将这条路径取出来存入结果 res 中,下面的部分是和一般的迭代中序写法有所不同的地方,由于中序遍历的特点,遍历到当前结点的时候,是有两种情况的,有可能此时是从左子结点跳回来的,此时正要去右子结点,则当前的结点值还是算在路径中的;也有可能当前是从右子结点跳回来的,并且此时要跳回上一个结点去,此时就要减去当前结点值,因为其已经不属于路径中的结点了。为了区分这两种情况,这里使用一个额外指针 pre 来指向前一个结点,如果右子结点存在且不等于 pre,直接将指针移到右子结点,反之更新 pre 为 cur,cur 重置为空,val 减去当前结点,st 删掉最后一个结点,参见代码如下:

 

解法二:

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> res;
        vector<TreeNode*> st;
        TreeNode *cur = root, *pre = nullptr;
        int val = 0;
        while (cur || !st.empty()) {
            while (cur) {
                st.push_back(cur);
                val += cur->val;
                cur = cur->left;
            }
            cur = st.back(); 
            if (!cur->left && !cur->right && val == sum) {
                vector<int> v;
                for (auto &a : st) v.push_back(a->val);
                res.push_back(v);
            }
            if (cur->right && cur->right != pre) {
                cur = cur->right;
            } else {
                pre = cur;
                val -= cur->val;
                st.pop_back();
                cur = nullptr;
            }
        }
        return res;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/113

 

类似题目:

Path Sum

Path Sum IV 

Path Sum III  

Binary Tree Maximum Path Sum

Sum Root to Leaf Numbers

Binary Tree Preorder Traversal

Binary Tree Paths

 

参考资料:

https://leetcode.com/problems/path-sum-ii/

https://leetcode.com/problems/path-sum-ii/discuss/36685/12ms-11-lines-C%2B%2B-Solution

https://leetcode.com/problems/path-sum-ii/discuss/36695/Java-Solution%3A-iterative-and-recursive

https://leetcode.com/problems/path-sum-ii/discuss/36683/DFS-with-one-LinkedList-accepted-java-solution

 

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2014-10-22 06:45  Grandyang  阅读(16404)  评论(6编辑  收藏  举报
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