[leetcode]Path Sum&&Path Sum II（剑指offer面试题25）

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题目描述：

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.

题目一是判断是否有一条从根到叶的路径使的路径上的节点值加起来等于给定的值。

代码如下：

/** * 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 false;        }        int curSum=0;        vector<vector<int>> result;        vector<int> path;        find(root,curSum,sum,path,result);        if(result.size()>0){            return true;        }        else{            return false;        }    }    void find(TreeNode *root, int curSum,int sum,vector<int> &path,vector<vector<int>> &result){        path.push_back(root->val);        curSum+=root->val;        bool isleaf=root->left==NULL&&root->right==NULL;        if(curSum==sum&&isleaf){            result.push_back(path);        }        if(root->left){            find(root->left,curSum,sum,path,result);        }        if(root->right){            find(root->right,curSum,sum,path,result);        }        path.pop_back();    }};

第二题是要打印路径思路一致

代码如下

/** * 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>> paths;        if(root==NULL){            return paths;        }        vector<int> result;        int curSum=0;        findPath(root,sum,curSum,result,paths);        return paths;    }    void findPath(TreeNode *root,int sum,int curSum,vector<int> &result,vector<vector<int>> &paths){        result.push_back(root->val);        curSum+=root->val;        bool isleaf=root->left==NULL&&root->right==NULL;        if(sum==curSum&&isleaf){            paths.push_back(result);        }        if(root->left!=NULL){            findPath(root->left,sum,curSum,result,paths);        }        if(root->right!=NULL){            findPath(root->right,sum,curSum,result,paths);        }        result.pop_back();    }};

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