65.二叉树的所有路径

给你一个二叉树的根节点 root ，按 任意顺序 ，返回所有从根节点到叶子节点的路径。

叶子节点 是指没有子节点的节点。

 

/*解题思路：回溯

1.递归参数：根节点
2.终止条件：节点无左右孩子*/

class Solution {
    List<String> res = new ArrayList<>();
    LinkedList<String> path = new LinkedList<>(); // 路径
    public List<String> binaryTreePaths(TreeNode root) {
        dfs(root);
        return res;
    }
    void dfs(TreeNode root) {
        path.add(String.valueOf(root.val));
        // 终止条件，无左右孩子
        if(root.left == null && root.right == null) {
            res.add(String.join("->", path));
            return;
        }
        if(root.left != null) {
            dfs(root.left);
            path.removeLast(); // 回溯
        } 
        if(root.right != null) {
            dfs(root.right);
            path.removeLast();
        }
    }
}
//解题思路：使用栈模拟递归
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    List<String> res = new ArrayList<>();
    // 栈模拟
    public List<String> binaryTreePaths(TreeNode root) {
        if(root == null) return res;
        Stack<Object> stack = new Stack<>(); // 栈      
        stack.push(root); stack.push(String.valueOf(root.val)); // 节点，当前路径
        while(!stack.isEmpty()) {
            String path = (String)stack.pop();
            TreeNode node = (TreeNode)stack.pop();
            if(node.left == null && node.right == null) res.add(path); // 递归终止条件
            if(node.right != null) {
                stack.push(node.right);
                stack.push(path + "->" + node.right.val);
            }
            if(node.left != null) {
                stack.push(node.left);
                stack.push(path + "->" + node.left.val);
            }
        }
        return res;    
    }
}