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"]
实现：
/**
* 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:
vector<string> binaryTreePaths(TreeNode* root) {
vector<string> vs;
if (root == NULL) return vs;
if (root->left == NULL && root->right == NULL) {
char buff[124];
sprintf(buff, "%d", root->val);
vs.push_back(buff);
return vs;
}
if (root->left) {
vector<string> lefts = binaryTreePaths(root->left);
for (int i = 0; i < lefts.size(); i++) {
char *buff = new char[lefts[i].size()+56];
sprintf(buff, "%d->%s", root->val, lefts[i].c_str());
vs.push_back(buff);
delete[] buff;
}
}
if (root->right) {
vector<string> rights = binaryTreePaths(root->right);
for (int i = 0; i < rights.size(); i++) {
char *buff = new char[rights[i].size()+56];
sprintf(buff, "%d->%s", root->val, rights[i].c_str());
vs.push_back(buff);
delete[] buff;
}
}
return vs;
}
};
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