题目描述

英文

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

中文

给定一个二叉树,返回所有从根节点到叶子节点的路径。

说明: 叶子节点是指没有子节点的节点。

示例:

输入:

   1
 /   \
2     3
 \
  5

输出: ["1->2->5", "1->3"]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

解法一

思路

这题可以用递归的方法来做,从根结点开始,对其左右子结点进行递归,递归的终止条件就是碰到叶子结点。每一层递归的输入是什么呢?结合题意,就是上层递归的输入加上上层结点的值了;而对于根结点,其输入就是空字符串。

以示例为例:

  • 第一次递归:输入'',输出'1'
  • 第二次递归:输入'1',左子结点输出'1->2',右子结点输出'1->3'(右子结点为叶子结点,停止递归)。
  • 第三次递归:输入'1->2',左子结点的右子结点输出'1->2->5'(左子结点的右子结点为叶子结点,停止递归)。
  • 输出:'1->3''1->2->5'

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """

        def paths(root, path, result):
            if root.left is None and root.right is None:
                result.append(path + str(root.val))
            if root.left is not None:
                paths(root.left, path + str(root.val) + '->', result)
            if root.right is not None:
                paths(root.right, path + str(root.val) + '->', result)

        result = []
        if root is not None:
            paths(root, '', result)
        return result

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        class Utils {
            public void paths(TreeNode root, String path, List<String> result) {
                if (root.left == null && root.right == null) {
                    result.add(path + root.val);
                }
                if (root.left != null) {
                    paths(root.left, path + root.val + "->", result);
                }
                if (root.right != null) {
                    paths(root.right, path + root.val + "->", result);
                }
            }
        }
        Utils utils = new Utils();
        List<String> result = new ArrayList<>();
        if (root != null) {
            utils.paths(root, "", result);
        }
        return result;
    }
}

C++

/**
 * 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) {
        class Utils {
        public:
            void paths(TreeNode *root, string path, vector<string> &result) {
                if (root->left == nullptr && root->right == nullptr) {
                    result.push_back(path + to_string(root->val));
                }
                if (root->left != nullptr) {
                    paths(root->left, path + to_string(root->val) + "->", result);
                }
                if (root->right != nullptr) {
                    paths(root->right, path + to_string(root->val) + "->", result);
                }
            }
        };
        vector<string> result;
        if (root != nullptr) {
            Utils utils;
            utils.paths(root, "", result);
        }
        return result;
    }
};