LeetCode_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.



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.

　　

/**

 * Definition for binary tree

 * 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) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

         queue<TreeNode *> myqueue;

        if(root == NULL) return false;

        myqueue.push(root) ;

        bool flag = false ;

        while(!myqueue.empty())

        {

          TreeNode * tp = myqueue.front();

          myqueue.pop();

          

          if(tp->left){

                tp->left->val +=tp->val ;

                myqueue.push(tp->left) ;

          }

          if(tp->right){

                tp->right->val +=tp->val ;

                myqueue.push(tp->right) ;

    

          }

          if(NULL == tp->left && NULL == tp->right )

          {

            if(tp->val == sum ) {

            flag = true ;

            break;

            }   

          }

         

        }

        

        return flag ;

    }

};

 DFS 解法：

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

public:

   bool DFS(TreeNode * root, int sum){    

        sum += root->val ;

        if(NULL == root->left && root->right == NULL){

            return sum == target;

        }

    

        if(root->left && DFS(root->left,sum)) return true;

        if(root->right && DFS(root->right,sum) ) return true;

        return false;

    }

    bool hasPathSum(TreeNode *root, int sum) {

        // Start typing your C/C++ solution below

        // DO NOT write int main() function

       target = sum;

       if(root == NULL) return false;

       return DFS(root, 0);

    }

    

private :

int target;

};

注意递归的退出条件： 最后的节点必须是叶子节点