LeetCode之“树”：Path Sum && Path Sum II

Path Sum

　　题目链接

　　题目要求：

　　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.

　　这道题采用深度优先搜索的方法就可以了，具体程序（12ms）如下：

/**
 * 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 false;
        return hasPathSumSub(root, 0, sum);
    }
    
    bool hasPathSumSub(TreeNode *tree, int subSum, int sum)
    {
        if(!tree)
            return false;
            
        subSum += tree->val;
        if(!tree->left && !tree->right && subSum == sum)
            return true;
        
        return hasPathSumSub(tree->left, subSum, sum) || hasPathSumSub(tree->right, subSum, sum);
    }
};

Path Sum II

　　题目链接

　　题目要求：

　　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]
]

　　这题与上题类似。具体程序（24ms）如下：

/**
 * 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:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> rVec;
        if(!root)
            return rVec;
        
        vector<int> vec;
        hasPathSumSub(root, vec, rVec, 0, sum);
        return rVec;
    }
    
    void hasPathSumSub(TreeNode *tree, vector<int> vec, vector<vector<int>>& rVec, int subSum, int sum)
    {
        if(!tree)
            return;
        
        vec.push_back(tree->val);
        subSum += tree->val;
        if(!tree->left && !tree->right && subSum == sum)
            rVec.push_back(vec);
        
        hasPathSumSub(tree->left, vec, rVec, subSum, sum);
        hasPathSumSub(tree->right, vec, rVec, subSum, sum);
    }
};