112. 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.

依然是迭代

/**
 * 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) 

        { 

             if(root==NULL)

               return false;

             else if(root->left==NULL&&root->right==NULL&&root->val==sum)

               return   true;

            else

                return   hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);

         }

}；