112. Path Sum (Tree; DFS)
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
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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(!root) return false;
flag = false;
target = sum;
preOrder(root,);
return flag;
}
void preOrder(TreeNode* node,int sum){
sum = node->val + sum; //递归前,加上当前节点
if(node->left)
{
preOrder(node->left,sum);
}
if(node->right)
{
preOrder(node->right,sum);
}
if(!node->left && !node->right && sum == target) //递归结束条件:到了叶子节点
{
flag = true;
}
}
private:
bool flag;
int target;
};
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