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257. Binary Tree Paths

Description

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

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• /**
   * 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 = new ArrayList<>();
      private List<String> t = new ArrayList<>();
  
      public List<String> binaryTreePaths(TreeNode root) {
          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));
          } else {
              dfs(root.left);
              dfs(root.right);
          }
          t.remove(t.size() - 1);
      }
  }
  

• /**
   * 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> ans;
          vector<string> t;
          function<void(TreeNode*)> dfs = [&](TreeNode* root) {
              if (!root) {
                  return;
              }
              t.push_back(to_string(root->val));
              if (!root->left && !root->right) {
                  ans.push_back(join(t));
              } else {
                  dfs(root->left);
                  dfs(root->right);
              }
              t.pop_back();
          };
          dfs(root);
          return ans;
      }
  
      string join(vector<string>& t, string sep = "->") {
          string ans;
          for (int i = 0; i < t.size(); ++i) {
              if (i > 0) {
                  ans += sep;
              }
              ans += t[i];
          }
          return ans;
      }
  };
  

• # 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: Optional[TreeNode]) -> List[str]:
          def dfs(root: Optional[TreeNode]):
              if root is None:
                  return
              t.append(str(root.val))
              if root.left is None and root.right is None:
                  ans.append("->".join(t))
              else:
                  dfs(root.left)
                  dfs(root.right)
              t.pop()
  
          ans = []
          t = []
          dfs(root)
          return ans
  
  

• /**
   * Definition for a binary tree node.
   * type TreeNode struct {
   *     Val int
   *     Left *TreeNode
   *     Right *TreeNode
   * }
   */
  func binaryTreePaths(root *TreeNode) (ans []string) {
  	t := []string{}
  	var dfs func(*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, "->"))
  		} else {
  			dfs(root.Left)
  			dfs(root.Right)
  		}
  		t = t[:len(t)-1]
  	}
  	dfs(root)
  	return
  }
  

• /**
   * 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[] {
      const ans: string[] = [];
      const t: number[] = [];
      const dfs = (root: TreeNode | null) => {
          if (!root) {
              return;
          }
          t.push(root.val);
          if (!root.left && !root.right) {
              ans.push(t.join('->'));
          } else {
              dfs(root.left);
              dfs(root.right);
          }
          t.pop();
      };
      dfs(root);
      return ans;
  }
  
  

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