【Sum Root to Leaf Numbers】cpp

题目:

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

 

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

代码:

/**
 * 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:
        int sumNumbers(TreeNode* root) 
        {
            return Solution::sumFromRoot2Leaf(root, 0);
        }
        static int sumFromRoot2Leaf(TreeNode* root, int upperVal)
        {
            if ( !root ) return 0;
            if ( !root->left && !root->right ) return upperVal*10+root->val;
            return Solution::sumFromRoot2Leaf(root->left, upperVal*10+root->val) + 
                   Solution::sumFromRoot2Leaf(root->right, upperVal*10+root->val);
        }
};

tips:

一开始想自低向上算,后来发现自顶向下算,每次传入上一层的值比较科学。

========================================

第二次过这道题,一次AC了,沿用dfs写法。

/**
 * 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:
        int sumNumbers(TreeNode* root)
        {
            int ret = 0;
            if ( root ) Solution::dfs(root, 0, ret);
            return ret;
        }
        static void dfs(TreeNode* root, int sum, int& ret )
        {
            if ( !root->left && !root->right )
            {
                sum = sum*10 + root->val;
                ret += sum;
                return;
            }
            if ( root->left ) Solution::dfs(root->left, sum*10+root->val, ret);
            if ( root->right ) Solution::dfs(root->right, sum*10+root->val, ret);
        }
};

 

posted on 2015-05-18 11:23  承续缘  阅读(224)  评论(0编辑  收藏  举报

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