Path Sum 判断二叉树的和 DFS处理

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​​

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
 //DFS处理
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        
        return dfs(root,sum,0);
    }
    
    bool dfs(TreeNode *node,int sum,int cur)
    {
        if(node==NULL)
            return false;
        if(node->left==NULL&&node->right==NULL)
            return sum==cur+node->val;
        return dfs(node->left,sum,cur+node->val)||dfs(node->right,sum,cur+node->val);
    }
};