<LeetCode OJ> 112/113. Path Sum(I / II)

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.

分析：

写在看别人答案之前：

1，不知道如何判断不是pathsum的情况，因为无法根据某一条不是就返回fasle。如果是倒数好判断

以下是错误答案，只通过了69个案例，有朝一日再来看

/** * 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 getDfsSums (TreeNode* node, int pathsum) {            if(!node->left && !node->right )                 {                if(pathsum==dstsum)                      return true;                else                    return false;             }            if(node->left)                  getDfsSums(node->left, pathsum+node->val);              if(node->right)                  getDfsSums(node->right, pathsum+node->val);        }     bool hasPathSum(TreeNode* root, int sum) {        if(!root)               return false;        dstsum=sum;        return getDfsSums(root, root->val);      }    int dstsum;};

别人的答案：

原来用或运算来判断有true的路劲

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; *///别人的答案：（我的分析）分多条路劲找，寻找结果进行或运算，只要有一条是真的，全部都是真的//这么简单的答案，艹...我有罪！class Solution {public:    bool hasPathSum(TreeNode* root, int sum) {        if(!root)            return false;        if(!root->left && !root->right ){            if(root->val==sum)                return true;            else                return false;        }        return hasPathSum(root->left, sum-(root->val)) || hasPathSum(root->right, sum-(root->val));    }};

学习和改写别人的答案：

/** * 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 IsPathSum (TreeNode* node, int pathsum) {  //此时的pathsum并未机上当前node的节点值（如果不为NULL的话）        if(!node)              return false;          if(!node->left && !node->right ){              if(pathsum+node->val==dstsum)  //如果加上当前节点就是sum，结果已出                return true;              else                  return false;          }                   return IsPathSum(node->left, pathsum+node->val)||IsPathSum(node->right, pathsum+node->val); //直接在函数中加上当前节点的值     }       bool hasPathSum(TreeNode* root, int sum) {        if(!root)              return false;          dstsum=sum;            return IsPathSum(root,0);      }    int dstsum;};

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]]

分析：
深度优先搜索，利用递归保存中间的计算结果，递归到叶子即刻判断和是否为目标sum

/** * 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 getDfsSums(vector<int> ans, TreeNode* node, int pathsum) {              ans.push_back(node->val);            pathsum+=node->val;                        if(!node->left && !node->right)  {                    if(pathsum==dstsum) result.push_back(ans);                return;              }                       if(node->left)                  getDfsSums(ans, node->left, pathsum);              if(node->right)                  getDfsSums(ans, node->right, pathsum);        }     vector<vector<int>> pathSum(TreeNode* root, int sum) {        if(!root)               return result;        vector<int> ans;        dstsum=sum;        getDfsSums(ans, root, 0);          return result;         }    int dstsum;    vector< vector<int> > result;  };

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原文地址：http://blog.csdn.net/ebowtang/article/details/50495606

原作者博客：http://blog.csdn.net/ebowtang