C++实现LeetCode(113.二叉树路径之和之二)

[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;
    }
};

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