[LeetCode] Sum Root to Leaf Numbers

最新推荐文章于 2020-03-06 22:26:33 发布

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最新推荐文章于 2020-03-06 22:26:33 发布

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本文介绍了一种解决二叉树中所有从根节点到叶子节点路径数字之和的方法。通过深度优先搜索（DFS）遍历二叉树，将每个路径上的数字连接起来形成整数，并计算所有路径形成的整数之和。

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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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Tree Depth-first Search

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;
        }
};