leetcode 437. Path Sum III | 递归!!

Description

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

Discuss

查看了几种方案, 递归的方式最简单, 其次运用leetcode 1 Two sum类似的简化方法.

双递归

• Java

public class Solution {
    public int pathSum(TreeNode root, int sum) {
        if(root == null)
            return 0;
        return findPath(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }
    
    public int findPath(TreeNode root, int sum){
        int res = 0;
        if(root == null)
            return res;
        if(sum == root.val)
            res++;
        res += findPath(root.left, sum - root.val);
        res += findPath(root.right, sum - root.val);
        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);
    }
private:
    int sumUp(TreeNode* root, int pre, int& sum){
        if(!root) return 0;
        int current = pre + root->val;
        return (current == sum) + sumUp(root->left, current, sum) + sumUp(root->right, current, sum);
    }
};

• 上述两段代码主要区别在于sumup的写法, 都有借鉴之处, 总结后自己代码如下所示

    // 第一种方式, 比较直观
    int sumup(TreeNode *root, int sum) {
        int res = 0;
        if (sum == root->val) ++res;
        if (root->left) res += sumup(root->left, sum - root->val);
        if (root->right) res += sumup(root->right, sum - root->val);
        return res;
    }
    // 第二种方式,比较简洁,更"递归范儿"
    int sumup(TreeNode *root, int sum) {
        if (!root) return 0;
        return (sum == root->val) + sumup(root->left, sum - root->val) + sumup(root->right, sum - root->val);
    }

待续

Reference

• leetcode 437. Path Sum III
• cplusplus unordered_map:find

Searches the container for an element with k as key and returns an iterator to it if found, otherwise it returns an iterator to unordered_map::end (the element past the end of the container).