LeetCode | Path Sum
2015-04-06 15:55
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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
return true, as there exist a root-to-leaf path
思路:本题意思是在二叉树中是否存在一条路径,该路径上的元素值总和为已知给出的sum。
因此,实际上可以将每层的结点值做一个转化,例如上述例子中第二层的结点值为4,8,那么到达第二层时,实际路径上的元素值和是5+4,5+8。即9,13.
所以,可以采用这种方法,逐层地下去,直到叶子节点。判断叶子结点值是否与sum相等即可。在程序中可以用递归的思路实现。
class Solution {
public:
bool hasPathSum(TreeNode *root, int sum)
{
if(!root)
return false;
else
return PathSum(root,0,sum);
}
bool PathSum(TreeNode* p, int total, int sum)
{
if(!p->left && !p->right)
return sum == total + p->val;
else if(p->left && !p->right)
return PathSum(p->left, total+p->val, sum);
else if(p->right && !p->left)
return PathSum(p->right, total+p->val, sum);
else
return (PathSum(p->left, total+p->val, sum) || PathSum(p->right, total+p->val, 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->2which sum is 22.
思路:本题意思是在二叉树中是否存在一条路径,该路径上的元素值总和为已知给出的sum。
因此,实际上可以将每层的结点值做一个转化,例如上述例子中第二层的结点值为4,8,那么到达第二层时,实际路径上的元素值和是5+4,5+8。即9,13.
所以,可以采用这种方法,逐层地下去,直到叶子节点。判断叶子结点值是否与sum相等即可。在程序中可以用递归的思路实现。
class Solution {
public:
bool hasPathSum(TreeNode *root, int sum)
{
if(!root)
return false;
else
return PathSum(root,0,sum);
}
bool PathSum(TreeNode* p, int total, int sum)
{
if(!p->left && !p->right)
return sum == total + p->val;
else if(p->left && !p->right)
return PathSum(p->left, total+p->val, sum);
else if(p->right && !p->left)
return PathSum(p->right, total+p->val, sum);
else
return (PathSum(p->left, total+p->val, sum) || PathSum(p->right, total+p->val, sum));
}
};
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