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.
思路:本題意思是在二叉樹中是否存在一條路徑,該路徑上的元素值總和爲已知給出的sum。
因此,實際上可以將每層的結點值做一個轉化,例如上述例子中第二層的結點值爲4,8,那麼到達第二層時,實際路徑上的元素值和是5+4,5+8。即9,13.
所以,可以採用這種方法,逐層地下去,直到葉子節點。判斷葉子結點值是否與sum相等即可。在程序中可以用遞歸的思路實現。
class Solution {
public:
bool hasPathSum(TreeNode *root, int sum)
{
if(!root)
return false;
else
return PathSum(root,0,sum);
}
bool PathSum(TreeNode* p, int total, int sum)
{
if(!p->left && !p->right)
return sum == total + p->val;
else if(p->left && !p->right)
return PathSum(p->left, total+p->val, sum);
else if(p->right && !p->left)
return PathSum(p->right, total+p->val, sum);
else
return (PathSum(p->left, total+p->val, sum) || PathSum(p->right, total+p->val, sum));
}
};