Question

Formatted question description: https://leetcode.ca/all/257.html

 257	Binary Tree Paths

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

 For example, given the following binary tree:

      1
    /  \
   2    3
    \
     5
 All root-to-leaf paths are:

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

 @tag-tree

Algorithm

In the recursive function, when a leaf node is encountered, there is no left or right child node, then a complete path has been formed at this time, add the current leaf node and save it in the result, and then backtrack.

Code

Java

import java.util.ArrayList;
import java.util.List;

public class Binary_Tree_Paths {

    public static void main (String[] args) {

        Binary_Tree_Paths out = new Binary_Tree_Paths();
        Solution s = out.new Solution();

        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.right = new TreeNode(5);

        System.out.println(s.binaryTreePaths(root));


        Solution_iteration si = out.new Solution_iteration();
        System.out.println(si.binaryTreePaths(root));
    }

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */

    class Solution_iteration {

        public List<Integer> binaryTreePaths(TreeNode root) {

            if (root == null) {
                return null;
            }

            Binary_Tree_Preorder_Traversal preOderIteration = new Binary_Tree_Preorder_Traversal();

            Binary_Tree_Preorder_Traversal.Solution preOrderSolution = preOderIteration.new Solution();

            List<Integer> result = preOrderSolution.preorderTraversal(root);

            return result;
        }
    }

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public List<String> binaryTreePaths(TreeNode root) {
            List<String> result = new ArrayList<String>();

            if(root == null) {
                return result;
            }

            helper(new String(), root, result);

            return result;
        }

        public void helper(String current, TreeNode root, List<String> result) {
            if(root.left == null && root.right == null) {
                result.add(current + root.val);
            }

            if(root.left != null) {
                helper(current + root.val + "->", root.left, result);
            }

            if(root.right != null) {
                helper(current + root.val + "->", root.right, result);
            }
        }

    }
}

// OJ: https://leetcode.com/problems/binary-tree-paths/
// Time: O(NH)
// Space: O(H^2)
class Solution {
private:
    vector<string> v;
    void preorder(TreeNode *root, string path) {
        path += to_string(root->val);
        if (!root->left && !root->right) {
            v.push_back(path);
            return;
        }
        path += "->";
        if (root->left) preorder(root->left, path);
        if (root->right) preorder(root->right, path);
    }
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        if (!root) return {};
        preorder(root, "");
        return v;
    }
};

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

class Solution:
  # @param {TreeNode} root
  # @return {string[]}
  def binaryTreePaths(self, root):
    def helper(root, path, res):
      if root:
        path.append(str(root.val))
        left = helper(root.left, path, res)
        right = helper(root.right, path, res)
        if not left and not right:
          res.append("->".join(path))
        path.pop()
        return True

    res = []
    helper(root, [], res)
    return res

257 - Binary Tree Paths

Question

Algorithm

Code

All Problems

All Solutions