【LeetCode & 劍指offer 刷題筆記】目錄（持續更新中...）

112. Path Sum

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. Note:  A leaf is a node with no children. 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.   /**  * Definition for a binary tree node.  * struct TreeNode {  *     int val;  *     TreeNode *left;  *     TreeNode *right;  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}  * };  */ /* 只要求返回true或false,因此不需要記錄路徑 */ class Solution { public :     bool hasPathSum ( TreeNode * root , int sum )     {         if ( root == nullptr ) return false ;         if ( root -> left == nullptr && root -> right == nullptr ) //葉子結點             return sum == root -> val ;                 int newsum = sum - root -> val ;         return hasPathSum ( root -> left , newsum ) || hasPathSum ( root -> right , newsum );     } };   113 .   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. Note:  A leaf is a node with no children. Example: Given the below binary tree and   sum = 22 ,

 

Return: [ [5,4,11,2], [5,8,4,5] ] /**  * Definition for a binary tree node.  * struct TreeNode {  *     int val;  *     TreeNode *left;  *     TreeNode *right;  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}  * };  */ /* 求所有和等於某數的路徑 */ #include <numeric> //算容器類元素和可用accumulate函式 class Solution { public :     vector < vector < int >> pathSum ( TreeNode * root , int sum )     {         vector < vector < int >> result ;         vector < int > path ;         path_sum ( root , path , result , sum );         return result ;     } private :     void path_sum ( TreeNode * root , vector < int >& path , vector < vector < int >>& result , int gap )     {         if ( root == nullptr )              return ; //遞迴出口         else                path . push_back ( root -> val ); // 儲存結點元素到path                 if ( root -> left == nullptr && root -> right == nullptr ) //葉子結點時push path到結果向量中         {             if ( gap == root -> val ) result . push_back ( path ); //如果該path和為sum則push到結果向量中（這裡用sum累減路徑上的元素，得到gap與路徑上最後一個元素比較，節省時間，如果得到path再accumulate，則會造成不同路徑間的重複計算）            // return; //遞迴出口，到葉結點後退出，（不能寫這句，還需執行到結尾進行pop）         }                 path_sum ( root -> left , path , result , gap - root -> val ); //沿深度方向遍歷         path_sum ( root -> right , path , result , gap - root -> val );         path . pop_back (); // 刪除最後一個元素 ，騰出空間(本函式中只push了一次，故只需pop一次)                     } };