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987. Vertical Order Traversal of a Binary Tree

  • Time: $O(n\log n/k) = O(n\log n)$, where $k = \text{width(tree)}$
  • Space: $O(n)$
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class Solution {
 public:
  vector<vector<int>> verticalTraversal(TreeNode* root) {
    vector<vector<int>> ans;
    map<int, multiset<pair<int, int>>> xToSortedPairs;  // {x: {(-y, val)}}

    dfs(root, 0, 0, xToSortedPairs);

    for (const auto& [_, pairs] : xToSortedPairs) {
      vector<int> vals;
      for (const pair<int, int>& pair : pairs)
        vals.push_back(pair.second);
      ans.push_back(vals);
    }

    return ans;
  }

 private:
  void dfs(TreeNode* root, int x, int y,
           map<int, multiset<pair<int, int>>>& xToSortedPairs) {
    if (root == nullptr)
      return;
    xToSortedPairs[x].emplace(y, root->val);
    dfs(root->left, x - 1, y + 1, xToSortedPairs);
    dfs(root->right, x + 1, y + 1, xToSortedPairs);
  }
};

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class Solution {
  public List<List<Integer>> verticalTraversal(TreeNode root) {
    List<List<Integer>> ans = new ArrayList<>();
    TreeMap<Integer, List<int[]>> xToSortedPairs = new TreeMap<>(); // {x: {(-y, val)}}

    dfs(root, 0, 0, xToSortedPairs);

    for (List<int[]> pairs : xToSortedPairs.values()) {
      Collections.sort(
          pairs,
          (a, b) -> a[0] == b[0] ? Integer.compare(a[1], b[1]) : Integer.compare(a[0], b[0]));
      List<Integer> vals = new ArrayList<>();
      for (int[] pair : pairs)
        vals.add(pair[1]);
      ans.add(vals);
    }

    return ans;
  }

  private void dfs(TreeNode root, int x, int y, TreeMap<Integer, List<int[]>> xToSortedPairs) {
    if (root == null)
      return;
    xToSortedPairs.putIfAbsent(x, new ArrayList<>());
    xToSortedPairs.get(x).add(new int[] {y, root.val});
    dfs(root.left, x - 1, y + 1, xToSortedPairs);
    dfs(root.right, x + 1, y + 1, xToSortedPairs);
  }
}

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class Solution:
  def verticalTraversal(self, root: TreeNode | None) -> list[list[int]]:
    ans = []
    xToNodes = collections.defaultdict(list)

    def dfs(node: TreeNode | None, x: int, y: int) -> None:
      if not node:
        return
      xToNodes[x].append((-y, node.val))
      dfs(node.left, x - 1, y - 1)
      dfs(node.right, x + 1, y - 1)

    dfs(root, 0, 0)

    for _, nodes in sorted(xToNodes.items(), key=lambda x: x[0]):
      ans.append([val for _, val in sorted(nodes)])

    return ans