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: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.
#include <iostream>
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
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;
}
if (!root->left && !root->right && root->val == sum)
{
return true;
}
int SumChild = sum - root->val;
if (hasPathSum(root->left, SumChild))
{
return true;
}
if (hasPathSum(root->right, SumChild))
{
return true;
}
return false;
}
};
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原文地址：http://blog.csdn.net/sheng_ai/article/details/46693781
