[LeetCode] 437. Path Sum III 路径和 III

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

给定一个二叉树，求从某一节点开始的路径的和等于给定值，不必从根节点开始，可从二叉树的任意一个节点开始，节点值有正有负。

解法：递归。

Python：

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        """
        def pathSumHelper(root, curr, sum, lookup):
            if root is None:
                return 0
            curr += root.val
            result = lookup[curr-sum] if curr-sum in lookup else 0
            lookup[curr] += 1
            result += pathSumHelper(root.left, curr, sum, lookup) + 
                      pathSumHelper(root.right, curr, sum, lookup)
            lookup[curr] -= 1
            if lookup[curr] == 0:
                del lookup[curr]
            return result
        
        lookup = collections.defaultdict(int)
        lookup[0] = 1
        return pathSumHelper(root, 0, sum, lookup)

Python:

class Solution2(object):
    def pathSum(self, root, sum):

        def pathSumHelper(root, prev, sum):
            if root is None:
                return 0

            curr = prev + root.val;
            return int(curr == sum) + 
                   pathSumHelper(root.left, curr, sum) + 
                   pathSumHelper(root.right, curr, sum)

        if root is None:
            return 0

        return pathSumHelper(root, 0, sum) + 
               self.pathSum(root.left, sum) + 
               self.pathSum(root.right, sum)

C++:

class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        unordered_map<int, int> m;
        m[0] = 1;
        return helper(root, sum, 0, m);
    }
    int helper(TreeNode* node, int sum, int curSum, unordered_map<int, int>& m) {
        if (!node) return 0;
        curSum += node->val;
        int res = m[curSum - sum];
        ++m[curSum];
        res += helper(node->left, sum, curSum, m) + helper(node->right, sum, curSum, m);
        --m[curSum];
        return res;
    }
};　　

C++:

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);
    }
};

　　

　　

　　

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