[LeetCode] Sum Root to Leaf Numbers

原文链接： http://www.cnblogs.com/diegodu/p/4424640.html

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.

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思路：典型dfs 深度搜索

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    private:
        int  m_sum;
        vector<int> m_vec;

        int intVec2Int(vector<int> nums)
        {
            int num = 0;
            for(int i = 0; i < nums.size(); i++)
            {
                num = num * 10 + nums[i];
            }
            return num;
        }

        void dfs(TreeNode* root)
        {
            if(root == NULL)
                return;

            m_vec.push_back(root->val);

            if(root->left == NULL && root->right == NULL)
            {
                m_sum += intVec2Int(m_vec);
                m_vec.pop_back();
                return;
            }

            if(root->left)
                dfs(root->left);
            if(root->right)
                dfs(root->right);
            m_vec.pop_back();

        }
    public:
        int sumNumbers(TreeNode *root)
        {
            m_vec.clear();
            m_sum = 0;
            dfs(root);
            return m_sum;
        }
};

转载于:https://www.cnblogs.com/diegodu/p/4424640.html