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.
判断二叉树是否有一条从根到叶的路径,使权值之和等于sum
/**
* 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 dfs(TreeNode* root,int now,int sum)
{
now+=root->val;
if(!root->left && !root->right) return now == sum;
bool l=false,r=false;
if(root->left) l=dfs(root->left,now,sum);
if(root->right) r=dfs(root->right,now,sum);
return l || r;
}
bool hasPathSum(TreeNode* root, int sum) {
if(!root) return false;
return dfs(root,0,sum);
}
};