Question

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

Given the root of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.

Example 1:

Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]

Example 2:

Input: root = [1]
Output: ["1"]

Constraints:

The number of nodes in the tree is in the range [1, 100].
-100 <= Node.val <= 100

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

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);
            }
        }

    }
}

############

/**
 * 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 {
    private List<String> ans;
    private List<String> t;

    public List<String> binaryTreePaths(TreeNode root) {
        ans = new ArrayList<>();
        t = new ArrayList<>();
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        t.add(root.val + "");
        if (root.left == null && root.right == null) {
            ans.add(String.join("->", t));
        }
        dfs(root.left);
        dfs(root.right);
        t.remove(t.size() - 1);
    }
}

// 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, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        def dfs(root):
            if root is None:
                return
            t.append(str(root.val))
            if root.left is None and root.right is None:
                ans.append('->'.join(t))
            dfs(root.left)
            dfs(root.right)
            t.pop()

        t = []
        ans = []
        dfs(root)
        return ans

############

# 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

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func binaryTreePaths(root *TreeNode) []string {
	var ans []string
	var t []string
	var dfs func(root *TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		t = append(t, strconv.Itoa(root.Val))
		if root.Left == nil && root.Right == nil {
			ans = append(ans, strings.Join(t, "->"))
		}
		dfs(root.Left)
		dfs(root.Right)
		t = t[:len(t)-1]
	}
	dfs(root)
	return ans
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function binaryTreePaths(root: TreeNode | null): string[] {
    let ans = [];
    let t = [];
    function dfs(root) {
        if (!root) return;
        t.push(String(root.val));
        if (!root.left && !root.right) ans.push(t.join('->'));
        dfs(root.left);
        dfs(root.right);
        t.pop();
    }
    dfs(root);
    return ans;
}

257 - Binary Tree Paths

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