Depth-first Search -- Leetcode problem113. Path Sum II

• 描述：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]
]

• 分析：这道题在path sum的基础上有改动，要求输出的是可能的所有情况的从根节点到叶子节点的值。
• 思路一；递归求解。(9ms)

class Solution {public:    vector<vector<int>> pathSum(TreeNode* root, int sum) {        vector<vector<int>> paths;        vector<int> path;        findPath(root, sum, 0, path, paths);        return paths;    }private:    void findPath(TreeNode* root, int sum, int add, vector<int> &path, vector<vector<int>> &paths) {        if (root == NULL) return;        add += root -> val;        path.push_back(root -> val);        if (root -> right == NULL && root -> left == NULL && add == sum) {            paths.push_back(path);        }        findPath(root -> left, sum, add, path, paths);        findPath(root -> right, sum, add, path, paths);        add -= root -> val;        path.pop_back();    }};

• 思路二：非递归DFS。使用三个栈来实现，其中一个栈用vector模仿。(9ms)

public:    vector<vector<int>> pathSum(TreeNode* root, int sum) {        vector<vector<int>> paths;        if (root == NULL) return paths;        stack<TreeNode*> my_stk;        vector<TreeNode*> history;        stack<int> result;        int temp = 0;        result.push(0);        my_stk.push(root);        while (!my_stk.empty()) {            TreeNode* top = my_stk.top();            my_stk.pop();            while (!history.empty()) {                TreeNode* pre = history.back();                if (pre -> left == top || pre -> right == top) {                    break;                }                history.pop_back();            }            temp = result.top();            result.pop();            temp += top -> val;            if (top -> left == NULL && top -> right == NULL && temp == sum) {                vector<int> path;                for (int i = 0; i < history.size(); i++) {                    path.push_back(history[i] -> val);                }                path.push_back(top -> val);                paths.push_back(path);            }            history.push_back(top);            if (top -> left) {                my_stk.push(top -> left);                result.push(temp);            }            if (top -> right) {                my_stk.push(top -> right);                result.push(temp);            }        }        return paths;    }};