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