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.

分析

给出一个二叉树和一个值，判断该树是否有一个根到叶的路径之和为该值。
需要依次递归记录路径上的值的和。然后加上叶子节点判断是否存在给定的和。
还可以不用另加一个数，将给定的和递减即可。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool hasSum(struct TreeNode* root, int sum,int pathsum)
{
    if(root->left==NULL&&root->right==NULL)
    {
        if(pathsum+root->val==sum)return true;
        else return false;
    }
    bool pathleft=false,pathright=false;
    if(root->left!=NULL)
        pathleft=hasSum(root->left,sum,pathsum+root->val);
    if(root->right!=NULL)
        pathright=hasSum(root->right,sum,pathsum+root->val);
    if(pathleft||pathright)
        return true;
    else
        return false;
}
bool hasPathSum(struct TreeNode* root, int sum) {
    if(root==NULL)return false;
    return hasSum(root,sum,0);
}