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.

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.

 

一开始被这题目坑了，因为叶子节点的左右孩子都是NULL  后捞改了一下代码  过了

/**
 * 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 hasPathSum1(TreeNode* root, int sum) {
        if (root == NULL){
            if (sum == 0) return true;
            return false;
        }
        if (root->left == NULL) return hasPathSum1(root->right,sum - root->val);
        else if (root->right == NULL) return hasPathSum1(root->left,sum - root->val);
        else return  (hasPathSum1(root->left,sum - root->val) || hasPathSum1(root->right,sum - root->val));
    }
    bool hasPathSum(TreeNode* root, int sum) {
        if (root == NULL) return false;
        return hasPathSum1(root, 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:
    bool hasPathSum(TreeNode* root, int sum) {
        
        if(!root) return false;
        if(sum == root->val && root->left==NULL &&root->right == NULL) return true;
         
        return hasPathSum(root->left, sum-root->val)||hasPathSum(root->right, sum-root->val);
        
    }
};