【Path Sum】cpp

题目：

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.

代码：

/**
 * 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) return false;
            int next = sum - root->val;
            if ( !root->left && !root->right ) return sum==root->val;
            if ( root->left && !root->right ) return Solution::hasPathSum(root->left, next);
            if ( !root->left && root->right ) return Solution::hasPathSum(root->right, next);
            return Solution::hasPathSum(root->left, next) || Solution::hasPathSum(root->right, next);
    }
};

tips：

一开始没理解好题意。

这个题要求必须走到某条path的叶子节点才算数，因此终止条件为走到叶子节点或者NULL。此外，root->left或者root->right不为NULL才往这个分支走。

============================================

第二次过这道题，终止条件是要走到叶子节点，但是参数传递写的有点儿啰嗦。

/**
 * 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)
        {
            bool find = false;
            if(root) Solution::pathSum(root, sum, 0, find);
            return find;
        }
        static void pathSum(TreeNode* root, int sum, int pathsum, bool& find)
        {
            if ( find ) return;
            if ( !root->left && !root->right ) 
            {
                if ( (pathsum+root->val)==sum )
                {
                    find = true;
                    return;
                }
            }
            if ( root->left ) Solution::pathSum(root->left, sum, pathsum+root->val, find);
            if ( root->right ) Solution::pathSum(root->right, sum, pathsum+root->val, find);
        }
};