LeetCode 437 Path Sum III

Problem:

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.

Summary:

找到二叉树中的子路径条数使得该子路径权值之和与给定和相等。

Analysis:

本题参考：http://www.cnblogs.com/grandyang/p/6007336.html

1. 最直观的思路：通过前序遍历的方式遍历二叉树，每遍历至一个节点，将当前节点的权值记录在数组中，并更新记录当前路径和的变量curSum，同时判断更新后的curSum与    sum是否相等，若相等则res加1。

此时的curSum为从根节点到当前节点的路径之和。至于这条路径上的子路径是否满足题意，则需将数组中的权值在curSum中一个个减掉，每剪掉一个，判断一次当前权值和是否与sum相等，若相等则res加1。此处需注意不能将数组中的权值完全减掉，最后应保留一个，因为全部去掉所得curSum为0，若题中所给的sum刚好为0，则判断相等，但不符合题意。

/**
 * 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*> vec;
        findPath(root, sum, 0, res, vec);
        
        return res;
    }
    
    void findPath(TreeNode* node, int sum, int curSum, int &res, vector<TreeNode*> &vec) {
        if (!node) return;
        curSum += node->val;
        vec.push_back(node);
        
        if (sum == curSum) res++;
        int tmp = curSum;
        for (int i = 0; i < vec.size() - 1; i++) {
            tmp -= vec[i]->val;
            if (tmp == sum) res++;
        }
        
        findPath(node->left, sum, curSum, res, vec);
        findPath(node->right, sum, curSum, res, vec);
        
        vec.pop_back();
    }
};

2. 优化算法，用hash表记录当前遍历路径中的子路径权值和对应出现的次数。

1. 若sum为从根节点到某x节点的路径权值和，则遍历至节点x时，当前的路径和curSum恰好与sum相等，则res = m[curSum - sum] = m[0] = 1;
2. 若sum为某段子路径权值和，如：x₁->x₂->x₃->x₄......中sum等于节点x₃与节点x₄的权值和，即sum = sum_x3+x4。则遍历至x₂时， m[curSum]++; 处已经记录了m[curSum] = m[sum_x1+x2] = 1,便利至x₄时curSum = sum_x1+x2+x3+x4,则res = m[curSum - sum] = m[sum_x1+x2+x3+x4 - sum_x3+x4] = m[sum_x1+x2] = 1。
   

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

3. 最简洁的方法，以每一个节点作为路径根节点进行前序遍历，查找每一条路径的权值和与sum是否相等。

class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        if (!root) return 0;
        int res = findPath(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
        return res;
    }
    
    int findPath(TreeNode* node, int curSum, int sum) {
        if (!node) return 0;
        curSum += node->val;
        return (curSum == sum) + findPath(node->left, curSum, sum) + findPath(node->right, curSum, sum);
    }
};