04
05/2015
[LeetCode] Path Sum
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:
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.
解题思路：
题意为判断从根到叶的路径上是否有和为给定的路径。递归即可。注意这种情况，若给出一个空树和0，应该返回false。
/**
* 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) {
if(root==NULL){
return false;
}
if(root->left==NULL&&root->right==NULL&&sum==root->val){
return true;
}
if(root->left!=NULL){
if(hasPathSum(root->left, sum - root->val)){
return true;
}
}
if(root->right!=NULL){
if(hasPathSum(root->right, sum - root->val)){
return true;
}
}
return false;
}
};二次刷题（2015-08-04）
还是犯了同样的错误，注意输入为空和0的情况应该返回false
/**
* 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) {
if(root==NULL){
return false;
}
if(root->left == NULL && root->right ==NULL){
return sum == root->val;
}
return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
}
};转载请注明：康瑞部落 » [LeetCode] Path Sum
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