LeetCode 113. Path Sum II

LeetCode 113. Path Sum II

Description:

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.

Example:

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]]

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分析：

如果采用栈去深度优先遍历解决，难度是比较大的，Discuss上面有一种解法采用了一个一维数组vector来存储一条路径，如果找到一条路径上的和等于sum的话，就将这个一维数组添加到二维数组中去。

同样，代码首先判断根节点是否为空，为空直接返回空二维数组。

不空时，将根节点加入到一维数组path中，然后判断左右子树是否为空，为空时判断根节点的数是否等于sum，若为真返回该path（只有根节点：一个节点一条路径）。

若左右子树不空时，递归左右子树。

注意最后有个路径回溯path.pop_back();

代码如下：

#include <iostream>#include <vector>using namespace std;/** * Definition for a binary tree node. */struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public:    vector<vector<int> > pathSum(TreeNode* root, int sum) {        vector<vector<int> > paths;        vector<int> path;        findPaths(root, sum, path, paths);        return paths;    }private:    void findPaths(TreeNode* node, int sum, vector<int>& path, vector<vector<int> >& paths) {        if (node == NULL) return;        path.push_back(node->val);        if (node->left == NULL && node->right == NULL && sum == node->val)            paths.push_back(path);        findPaths(node->left, sum - node->val, path, paths);        findPaths(node->right, sum - node->val, path, paths);        path.pop_back();    }};// 构造二叉树int TreeNodeCreate(TreeNode* &tree) {    int val;    cin >> val;    if (val < 0) // 小于0表示空节点        tree = NULL;    else {        tree = new TreeNode(val); // 创建根节点        tree->val = val;        TreeNodeCreate(tree->left); // 创建左子树        TreeNodeCreate(tree->right);// 创建右子树    }    return 0;}int main() {    Solution s;    TreeNode* tree;    TreeNodeCreate(tree);    int sum;    cin >> sum;    vector<vector<int> > res = s.pathSum(tree, sum);    for (int i = 0; i < res.size(); i++) {        for (int j = 0; j < res[i].size(); j++) {            cout << res[i][j] << " ";        }        cout << endl;    }    return 0;}/*input:54117-1 -12-1 -1-1813-1 -145-1 -11-1 -122output:5 4 11 25 8 4 5 */