<LeetCode OJ> 257. Binary Tree Paths

257. Binary Tree Paths

Total Accepted: 29282 Total Submissions: 113527 Difficulty: Easy

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

["1->2->5", "1->3"]

分析：

这个算法写的太精妙了，參考讨论区!

/**
 * 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:
    void getDfsPaths(vector<string>& result, TreeNode* node, string strpath) {
        if(!node->left && !node->right){//叶子
            result.push_back(strpath);
            return ;
        }
        if(node->left)
            getDfsPaths(result, node->left, strpath+"->"+to_string(node->left->val));
        if(node->right)
            getDfsPaths(result, node->right, strpath+"->"+to_string(node->right->val));
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> ret;
        if(!root) 
            return ret;
        
        getDfsPaths(ret, root, to_string(root->val));
        return ret;
    }
};

小结：

1。深度搜索应该立马条件反射,採用前序式遍历（假设用递归的话）

2，深度优先搜索应该立马联想到栈来实现迭代

3，递归具有保存变量信息的功能，有时候值得利用

联动第二十二题

【1】 22. Generate Parentheses。

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