Leetcoder-112-path Sum

Path Sum

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such this 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           /   /           /  4         /  \              7    2      1

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

Whether there is a sum from the root node to the leaf node equals the number given.

Recursive one

/** * 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:     void Solve (treenode* root,int sum,bool& f,int s) {        if (!root->left&&! Root->right) {            if (s==sum) f=true;            return;        }        if (root->left) solve (root->left,sum,f,s+root->left->val);        if (root->right) solve (root->right,sum,f,s+root->right->val);    }    BOOL Haspathsum (treenode* root, int sum) {        bool f=false;        if (!root) return false;//is empty  then return false                solve (root,sum,f,root->val);        return f;    };

Recursive

/** * 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:    bool Haspathsum (treenode* root, int sum) {        return solve (root,sum,0);    }    BOOL Solve (treenode* root,int sum,int s) {        if (!root) return false;        if (!root->left&&!root->right)             return sum==s+root->val;                return solve (root->left,sum,s+root->val) | | Solve (root->right,sum,s+root->val);//existence can be used | |     };

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Leetcoder-112-path Sum