LeetCode——Path Sum III

LeetCode——Path Sum III

#437

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

10
/  \
5   -3
/ \    \
3   2   11
/ \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

这一题的目的是找出路径值之和等于给定值的个数。这一题第一个思路就是利用递归。

构造一个递归函数，对于树中的每个点，先判断根路径到这个节点的和是否为目标值，如果不是，则从上往下去掉一个节点的值，但要保留一个，否则为0，如果目标值为0，就会返回错误的值。从根节点开始分别递归左右子树。

C++

/**
* 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:
int pathSum(TreeNode* root, int sum) {
int res = 0;
vector<TreeNode*> out;
helper(root,sum,0,out,res);
return res;
}
void helper(TreeNode* Node,int sum,int cursum,vector<TreeNode*>& out,int& res) { //使用了引用作为参数
if(!Node)
return;
cursum += Node -> val;
out.push_back(Node);
if(cursum == sum)
res++;
int t = cursum;
for(int i = 0;i < out.size() - 1;i++) {
t -= out[i] -> val;
if(t == sum)
a28e

res++;
}
helper(Node -> left,sum,cursum,out,res);
helper(Node -> right,sum,cursum,out,res);
out.pop_back();
}
};

还有一种简洁的写法，也是利用了先序遍历，对于每个遍历到的节点进行处理，维护一个变量pre来记录之前路径之和，然后cur为pre加上当前节点值，如果cur等于sum，那么返回结果时要加1，然后对当前节点的左右子节点调用递归函数求解。

C++

/**
* 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:
int pathSum(TreeNode* root, int sum) {
if (!root) return 0;
return sumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
}
int sumUp(TreeNode* node, int pre, int& sum) {
if (!node) return 0;
int cur = pre + node->val;
return (cur == sum) + sumUp(node->left, cur, sum) + sumUp(node->right, cur, sum);
}
};