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, , 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, , 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);
     }
 };