原题如下:
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.
bool hasPathSum(TreeNode *root, int sum) {
if(root == NULL)
return false;
int curSum = 0;
return pathSum(root,curSum,sum) ;
}
bool pathSum(TreeNode *root,int curSum,int sum){
if(root == NULL && curSum == sum)
return true;
if(root == NULL)
return false;
curSum += root->val;
if(root->left != NULL && root->right == NULL && curSum ==sum)
return pathSum(root->left,curSum,sum);
if(root->right != NULL && root->left == NULL && curSum == sum)
return pathSum(root->right,curSum,sum);
return (pathSum(root->left,curSum,sum) || pathSum(root->right,curSum,sum));
}
在上述方法中,我是在遍历节点之后再求和,所以其递归结束的条件设定为节点为空且和刚好相等,但此时遇到了问题,那就是树{1,2}和为1时将会判断错误,其主要原因是左右子树不同时为空时判断出错,所以,在求和之后加入了两个if条件来进行判断,最后测试通过。
考虑到上述问题存在的原因是递归结束条件判断不够合理,所以我又在上述基础上进行了改进:先求和在遍历节点,但此时需要保证所遍历的节点不为空,代码如下:
bool hasPathSum2(TreeNode *root, int sum) {
if(root == NULL)
return false;
int curSum = root->val;
return pathSum2(root,curSum,sum) ;
}
bool pathSum2(TreeNode *root,int curSum,int sum){
if(root->left == NULL && root->right == NULL && curSum == sum)
return true;
bool b1 = false,b2 = false;
if(root->left != NULL){
int sum1 = curSum + root->left->val;
b1 = pathSum2(root->left,sum1,sum);
}
if(root->right != NULL){
int sum2 = curSum + root->right->val;
b2 = pathSum2(root->right,sum2,sum);
}
return b1 || b2;
}
改进后的代码比之前明晰了好多,所以在利用递归时,递归结束条件的判断很重要,一定要根据题意选择恰当的递归结束条件。