Leetcode---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.

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Have you met this question in a real interview?

这个题关键在于如何在递归中计算遍历的节点的和，一个技巧就是入口处加，出口处减，当然还有的诀窍是传值传递参数

首先看用全局变量或者引用传递的方法，用时20ms

int num=0;
bool hasPathSum(TreeNode *root, int sum){
    if(root==NULL)
        return false;
    num+=root->val;
    if(root->left==NULL && root->right==NULL){
        if(num==sum){
            num-=root->val;
            return true;
        }
        else{
            num-=root->val;
            return false;
        }
    }
    bool result= hasPathSum(root->left,sum)||hasPathSum(root->right,sum);
    num-=root->val;
    return result;
}

然后看值传递的方式：耗时80ms，明显慢一些，和传了三个参数关系不大，因为去掉成两个参数之后还是76ms

bool f(TreeNode *root, int& sum, int num){
    if(root==NULL)
        return false;
    num+=root->val;
    if(root->left==NULL && root->right==NULL){
        if(num==sum){
            return true;
        }
        else{
            return false;
        }
    }
    return f(root->left,sum,num)||f(root->right,sum,num);
}
bool hasPathSum(TreeNode *root, int sum){
    return f(root,sum,0);
}