<p>Given the <code>root</code> of a binary tree, return <em>all root-to-leaf paths in <strong>any order</strong></em>.</p>

<p>A <strong>leaf</strong> is a node with no children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/12/paths-tree.jpg" style="width: 207px; height: 293px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,5]
<strong>Output:</strong> [&quot;1-&gt;2-&gt;5&quot;,&quot;1-&gt;3&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1]
<strong>Output:</strong> [&quot;1&quot;]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 100]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> paths;
        if (!root)
            return paths;

        stack<pair<TreeNode*, string>> nodeStack;
        nodeStack.push(
            {root, to_string(root->val)}); // 初始化栈，存储根节点和初始路径

        while (!nodeStack.empty()) {
            auto [node, path] = nodeStack.top();
            nodeStack.pop();

            // 如果是叶子节点，将路径加入结果
            if (!node->left && !node->right) {
                paths.push_back(path);
            }

            // 如果存在右子节点，先将右子节点压入栈
            if (node->right) {
                nodeStack.push(
                    {node->right, path + "->" + to_string(node->right->val)});
            }

            // 如果存在左子节点，再将左子节点压入栈
            if (node->left) {
                nodeStack.push(
                    {node->left, path + "->" + to_string(node->left->val)});
            }
        }

        return paths;
    }
};
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