leetcode 112/113/437. Path Sum 1/2/3

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本文探讨了二叉树中寻找特定路径和的问题，包括单条路径求和判断、查找所有符合要求的路径以及统计任意路径组合达到指定和的数目。通过递归算法解决这些挑战，并提供详细的代码实现。

Path Sum I

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.

直接递归。

/**
 * 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:
    bool hasPathSum(TreeNode* root, int sum) 
    {
        if (!root)
            return 0;
        else if (!root->left && !root->right)
        {
            if (root->val == sum) return 1;
            else    return 0;          
        }
        else
            return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};

Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

也是直接递归

/**
 * 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:
    void helper(TreeNode* p, int sum, vector<vector<int>> & result, vector<int> m1)
    {
        if (!p)
            return;
        
        m1.push_back(p->val);
        if (!p->left && !p->right && p->val == sum)
            result.push_back(m1);

        helper(p->left, sum - p->val, result, m1);
        helper(p->right, sum - p->val, result, m1);
    }
    
    vector<vector<int>> pathSum(TreeNode* root, int sum) 
    {
        vector<vector<int>> result;
        vector<int> m1;
        helper(root, sum, result, m1);
        return result;  
    } 
};

Path Sum 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

直接用一个queue来记录沿路的路径，每一个点都可以作为开始。

/**
 * 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 target)
    {
        if (!root) return 0;
        ret = 0;
        queue<int> mp;
        helper(mp, root, target);
        return ret;
    }
    
    void helper(queue<int> que, TreeNode* root, int target)
    {
        int size = que.size();
        for (int i = 0; i < size; i++)
        {
            int key = que.front() + root->val;
            if (key == target)
                ret ++;
            que.pop();
            que.push(key);
        }
        if (root->val == target)    ret++;
        que.push(root->val);
        if (root->left) helper(que, root->left, target);
        if (root->right) helper(que, root->right, target);
        return;
    }
private:
    int ret;
};