leetCode(40):Path Sum 分类： leetCode 2015-07-17 16:46 125人阅读 评论(0) 收藏

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.

搜索支路的方法

/**
* 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 findPath(TreeNode* root, int nowSum,const int sum)
{
nowSum += root->val;//当前值

//如果已经是叶子结点，且和相等，则直接返回真
if (!root->left && !root->right && nowSum == sum)
return true;

bool found_left = false;
bool found_right = false;

if (root->left)	//左子树不为空，则在左子树中搜索
found_left = findPath(root->left, nowSum, sum);

if (!found_left && root->right)	//如果左子树没找到，则在右子树中搜索
found_right = findPath(root->right, nowSum, sum);

return found_left || found_right;//只要在一条支中上找到则返回真
}

bool hasPathSum(TreeNode* root, int sum) {
if (root == NULL)
return false;
findPath(root, 0, sum);
}
};