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1022. Sum of Root To Leaf Binary Numbers 👍

  • Time: $O(n)$
  • Space: $O(h)$
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class Solution {
 public:
  int sumRootToLeaf(TreeNode* root) {
    int ans = 0;
    dfs(root, 0, ans);
    return ans;
  }

 private:
  void dfs(TreeNode* root, int val, int& ans) {
    if (root == nullptr)
      return;
    val = val * 2 + root->val;
    if (root->left == nullptr && root->right == nullptr)
      ans += val;
    dfs(root->left, val, ans);
    dfs(root->right, val, ans);
  }
};

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class Solution {
  public int sumRootToLeaf(TreeNode root) {
    dfs(root, 0);
    return ans;
  }

  private int ans = 0;

  private void dfs(TreeNode root, int val) {
    if (root == null)
      return;
    val = val * 2 + root.val;
    if (root.left == null && root.right == null)
      ans += val;
    dfs(root.left, val);
    dfs(root.right, val);
  }
}

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class Solution:
  def sumRootToLeaf(self, root: Optional[TreeNode]) -> int:
    ans = 0

    def dfs(root: Optional[TreeNode], val: int) -> None:
      nonlocal ans
      if not root:
        return
      val = val * 2 + root.val
      if not root.left and not root.right:
        ans += val
      dfs(root.left, val)
      dfs(root.right, val)

    dfs(root, 0)
    return ans