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.
分析
给出一个二叉树和一个值，判断该树是否有一个根到叶的路径之和为该值。
需要依次递归记录路径上的值的和。然后加上叶子节点判断是否存在给定的和。
还可以不用另加一个数，将给定的和递减即可。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool hasSum(struct TreeNode* root, int sum,int pathsum)
{
if(root->left==NULL&&root->right==NULL)
{
if(pathsum+root->val==sum)return true;
else return false;
}
bool pathleft=false,pathright=false;
if(root->left!=NULL)
pathleft=hasSum(root->left,sum,pathsum+root->val);
if(root->right!=NULL)
pathright=hasSum(root->right,sum,pathsum+root->val);
if(pathleft||pathright)
return true;
else
return false;
}
bool hasPathSum(struct TreeNode* root, int sum) {
if(root==NULL)return false;
return hasSum(root,sum,0);
}
