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Binary Tree Paths.cpp
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Binary Tree Paths.cpp
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/*
Given a binary tree, return all root-to-leaf paths.
Link: http://www.lintcode.com/en/problem/binary-tree-paths/
Example:
Solution: update the path, and use path to construct string when we need to simple the
process on reduce path list.
Source: https://github.com/kamyu104/LintCode/blob/master/C%2B%2B/binary-tree-paths.cpp
*/
/**
* 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) {
vector<string> result;
vector<TreeNode *> path;
binaryTreePathsRecu(root, &path, &result);
return result;
}
void binaryTreePathsRecu(TreeNode *node, vector<TreeNode *> *path,
vector<string> *result) {
if (!node) {
return;
}
if (!node->left && !node->right) {
string ans = "";
for (const auto& n : *path) {
ans.append(to_string(n->val).append("->"));
}
// move is used to indicate that an object t may be "moved from",
// i.e. allowing the efficient transfer of resources from t to another object
// aims to change the content from left and right
result->emplace_back(move(ans.append(to_string(node->val))));
}
if (node->left) {
path->emplace_back(node);
binaryTreePathsRecu(node->left, path, result);
path->pop_back();
}
if (node->right) {
path->emplace_back(node);
binaryTreePathsRecu(node->right, path, result);
path->pop_back();
}
}
};