257BinaryTreePaths.org

[#C] 257 Binary Tree Paths - LeetCode

题目

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

思路：通过递归叠加每个节点

Java

/**
 * 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 {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> res = new ArrayList();
        if (root == null) return res;
     
        if (root.left == null && root.right == null) {
            res.add(String.valueOf(root.val));
            return res;
        }
     
        List<String> lefts = binaryTreePaths(root.left);
        for (int i = 0; i < lefts.size(); i++) {
            res.add(root.val + "->" + lefts.get(i));
        }
     
        List<String> rights = binaryTreePaths(root.right);
        for (int i = 0; i < rights.size(); i++) {
            res.add(root.val + "->" + rights.get(i));
        }
     
        return res;
    }
}

CPP

/**
 * 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> res;
     
        if (root == NULL) return res;
        if (root->left == NULL && root->right == NULL) {
            res.push_back(to_string(root->val));
            return res;
        }
     
        vector<string> lefts = binaryTreePaths(root->left);
        for (int i = 0; i < lefts.size(); i++) {
            res.push_back(to_string(root->val) + "->" + lefts[i]);
        }
     
        vector<string> rights = binaryTreePaths(root->right);
        for (int i = 0; i < rights.size(); i++) {
            res.push_back(to_string(root->val) + "->" + rights[i]);
        }
     
        return res;
    }
};