题目介绍

力扣257题：leetcode-cn.com/problems/bi…

方法一：递归 + 回溯

首先我们可以知道，找到了叶子节点，也就找到了一条路径，叶子节点的判断条件很简单，就是左子树跟右子树都为空，所以思路还是比较简单的，代码如下；

/**
 * 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> result = new ArrayList<>();
    public List<String> binaryTreePaths(TreeNode root) {
        if(root == null) {
            return result;
        }
        List<Integer> paths = new ArrayList<>();
        preOrder(root, paths);
        return result;
    }

    public void preOrder(TreeNode root, List<Integer> paths) {
        paths.add(root.val);
        //找到了叶子节点，即找到了一条路径
        if (root.left == null && root.right == null) {
            StringBuilder sb = new StringBuilder();
            for (Integer i : paths) {
                sb.append(i).append("->");
            }
            //去掉最后一个多余的->
            sb.delete(sb.length() - 2, sb.length());
            result.add(sb.toString());
            return;
        }
        if (root.left != null) {
            preOrder(root.left, paths);
            //回溯
            paths.remove(paths.size() - 1);
        }

        if (root.right != null) {
            preOrder(root.right, paths);
            //回溯
            paths.remove(paths.size() - 1);
        }   
    }
}

方法二：迭代

代码如下：

// 解法2
class Solution {
    /**
     * 迭代法
     */
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<>();
        if (root == null)
            return result;
        Stack<Object> stack = new Stack<>();
        // 节点和路径同时入栈
        stack.push(root);
        stack.push(root.val + "");
        while (!stack.isEmpty()) {
            // 节点和路径同时出栈
            String path = (String) stack.pop();
            TreeNode node = (TreeNode) stack.pop();
            // 若找到叶子节点
            if (node.left == null && node.right == null) {
                result.add(path);
            }
            //右子节点不为空
            if (node.right != null) {
                stack.push(node.right);
                stack.push(path + "->" + node.right.val);
            }
            //左子节点不为空
            if (node.left != null) {
                stack.push(node.left);
                stack.push(path + "->" + node.left.val);
            }
        }
        return result;
    }
}