标签:leetcode
Path Sum
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.
递归 one
/**
* 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:
void solve(TreeNode* root,int sum,bool& f,int s){
if(!root->left&&!root->right){
if(s==sum) f=true;
return;
}
if(root->left) solve(root->left,sum,f,s+root->left->val);
if(root->right) solve(root->right,sum,f,s+root->right->val);
}
bool hasPathSum(TreeNode* root, int sum) {
bool f=false;
if(!root) return false;// 为空 则返回false
solve(root,sum,f,root->val);
return f;
}
};
递归 two
/**
* 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 hasPathSum(TreeNode* root, int sum) {
return solve(root,sum,0);
}
bool solve(TreeNode* root,int sum,int s){
if(!root) return false;
if(!root->left&&!root->right)
return sum==s+root->val;
return solve(root->left,sum,s+root->val)||solve(root->right,sum,s+root->val);//存在即可 故用 ||
}
};版权声明:本文为博主原创文章,未经博主允许不得转载。
标签:leetcode
原文地址:http://blog.csdn.net/u014705854/article/details/46822123