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"]
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
//这个算法写的太精妙了，参考讨论区，别人家的!
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;
}
};注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！
原文地址：http://blog.csdn.net/ebowtang/article/details/50493936
原作者博客：http://blog.csdn.net/ebowtang
<LeetCode OJ> 257. Binary Tree Paths
原文：http://blog.csdn.net/ebowtang/article/details/50493936