lintcode 二叉樹的所有路徑

每日一練 加油

描述

給一棵二叉樹，找出從根節點到葉子節點的所有路徑。

樣例

給出下面這棵二叉樹：

1
/ \
2 3
\
5

所有根到葉子的路徑為：

[
“1->2->5”,
“1->3”
]

思路:

對於某一個根節點root，他的左右子樹都會分別產生2個路徑的集合vector left 和right 那麼只需要把這兩個集合裡所有的路徑的頭部新增一下root，再把兩個vector合併即可。
空結點和葉節點處理一下就完成了

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: the root of the binary tree
     * @return: all root-to-leaf paths
     */
    vector<string> binaryTreePaths(TreeNode * root) {
        // write your code here
        vector<string> res;
        if (!root)
            return res;
        if (root->left == NULL && root->right == NULL) 
            res.push_back(to_string(root->val));    
        string head = to_string(root->val);
        vector<string> left = binaryTreePaths(root->left),
                       right = binaryTreePaths(root->right);
        for (int i = 0; i < left.size(); i++)
            left[i] = head + "->" + left[i]; 
            
        for (int i = 0; i < right.size(); i++)
            right[i] = head + "->" + right[i]; 
        res.insert(res.end(), left.begin(), left.end());
        res.insert(res.end(), right.begin(), right.end());
        return res;
        
    }
};