Leetcode:Path Sum与Path Sum II

Path Sum：

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5             / \            4   8           /   / \          11  13  4         /  \      \        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

递归，实现代码：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool hasPathSum(TreeNode *root, int sum) {        if(root==NULL)            return 0;        if(root->left==NULL&&root->right==NULL)            return root->val==sum;        return hasPathSum(root->left,sum-root->val)||                   hasPathSum(root->right,sum-root->val);    }};

Path Sum 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]]

实现代码：

/** * Definition for binary tree * 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> > res;        vector<int> path;        findPath(root, sum, path, res);        return res;    }    void findPath(TreeNode* root, int sum, vector<int> path, vector<vector<int> > & res)    {        if(root==NULL)            return;        path.push_back(root->val);        if(sum == root->val && root->left==NULL && root->right==NULL)        {            res.push_back(path);            path.pop_back();            return;        }        if (root->left)            findPath(root->left, sum-root->val,path,res);        if (root->right)            findPath(root->right,sum-root->val,path,res);    path.pop_back ();    }};