Question

314. 二叉树的垂直遍历

English Version

题目描述

给你一个二叉树的根结点，返回其结点按 垂直方向（从上到下，逐列）遍历的结果。

如果两个结点在同一行和列，那么顺序则为 从左到右。

 

示例 1：

输入：root = [3,9,20,null,null,15,7]
输出：[[9],[3,15],[20],[7]]

示例 2：

输入：root = [3,9,8,4,0,1,7]
输出：[[4],[9],[3,0,1],[8],[7]]

示例 3：

输入：root = [3,9,8,4,0,1,7,null,null,null,2,5]
输出：[[4],[9,5],[3,0,1],[8,2],[7]]

 

提示：

• 树中结点的数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

解法

方法一：DFS

DFS 遍历二叉树，记录每个节点的值、深度，以及横向的偏移量。然后对所有节点按照横向偏移量从小到大排序，再按照深度从小到大排序，最后按照横向偏移量分组。

时间复杂度 $O(n\log \log n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。

# 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 verticalOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
    def dfs(root, depth, offset):
      if root is None:
        return
      d[offset].append((depth, root.val))
      dfs(root.left, depth + 1, offset - 1)
      dfs(root.right, depth + 1, offset + 1)

    d = defaultdict(list)
    dfs(root, 0, 0)
    ans = []
    for _, v in sorted(d.items()):
      v.sort(key=lambda x: x[0])
      ans.append([x[1] for x in v])
    return ans

/**
 * 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 TreeMap<Integer, List<int[]>> d = new TreeMap<>();

  public List<List<Integer>> verticalOrder(TreeNode root) {
    dfs(root, 0, 0);
    List<List<Integer>> ans = new ArrayList<>();
    for (var v : d.values()) {
      Collections.sort(v, (a, b) -> a[0] - b[0]);
      List<Integer> t = new ArrayList<>();
      for (var e : v) {
        t.add(e[1]);
      }
      ans.add(t);
    }
    return ans;
  }

  private void dfs(TreeNode root, int depth, int offset) {
    if (root == null) {
      return;
    }
    d.computeIfAbsent(offset, k -> new ArrayList<>()).add(new int[] {depth, root.val});
    dfs(root.left, depth + 1, offset - 1);
    dfs(root.right, depth + 1, offset + 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) {}
 * };
 */
using pii = pair<int, int>;

class Solution {
public:
  map<int, vector<pii>> d;

  vector<vector<int>> verticalOrder(TreeNode* root) {
    dfs(root, 0, 0);
    vector<vector<int>> ans;
    for (auto& [_, v] : d) {
      sort(v.begin(), v.end(), [&](pii& a, pii& b) {
        return a.first < b.first;
      });
      vector<int> t;
      for (auto& x : v) {
        t.push_back(x.second);
      }
      ans.push_back(t);
    }
    return ans;
  }

  void dfs(TreeNode* root, int depth, int offset) {
    if (!root) return;
    d[offset].push_back({depth, root->val});
    dfs(root->left, depth + 1, offset - 1);
    dfs(root->right, depth + 1, offset + 1);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func verticalOrder(root *TreeNode) [][]int {
  d := map[int][][]int{}
  var dfs func(*TreeNode, int, int)
  dfs = func(root *TreeNode, depth, offset int) {
    if root == nil {
      return
    }
    d[offset] = append(d[offset], []int{depth, root.Val})
    dfs(root.Left, depth+1, offset-1)
    dfs(root.Right, depth+1, offset+1)
  }
  dfs(root, 0, 0)
  idx := []int{}
  for i := range d {
    idx = append(idx, i)
  }
  sort.Ints(idx)
  ans := [][]int{}
  for _, i := range idx {
    v := d[i]
    sort.SliceStable(v, func(i, j int) bool { return v[i][0] < v[j][0] })
    t := []int{}
    for _, x := range v {
      t = append(t, x[1])
    }
    ans = append(ans, t)
  }
  return ans
}

方法二：BFS

本题较好的做法应该是 BFS，从上往下逐层进行遍历。

时间复杂度 $O(n\log n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的结点数。

# 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 verticalOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
    if root is None:
      return []
    q = deque([(root, 0)])
    d = defaultdict(list)
    while q:
      for _ in range(len(q)):
        root, offset = q.popleft()
        d[offset].append(root.val)
        if root.left:
          q.append((root.left, offset - 1))
        if root.right:
          q.append((root.right, offset + 1))
    return [v for _, v in sorted(d.items())]

/**
 * 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 {
  public List<List<Integer>> verticalOrder(TreeNode root) {
    List<List<Integer>> ans = new ArrayList<>();
    if (root == null) {
      return ans;
    }
    Deque<Pair<TreeNode, Integer>> q = new ArrayDeque<>();
    q.offer(new Pair<>(root, 0));
    TreeMap<Integer, List<Integer>> d = new TreeMap<>();
    while (!q.isEmpty()) {
      for (int n = q.size(); n > 0; --n) {
        var p = q.pollFirst();
        root = p.getKey();
        int offset = p.getValue();
        d.computeIfAbsent(offset, k -> new ArrayList()).add(root.val);
        if (root.left != null) {
          q.offer(new Pair<>(root.left, offset - 1));
        }
        if (root.right != null) {
          q.offer(new Pair<>(root.right, offset + 1));
        }
      }
    }
    return new ArrayList<>(d.values());
  }
}

/**
 * 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<vector<int>> verticalOrder(TreeNode* root) {
    vector<vector<int>> ans;
    if (!root) return ans;
    map<int, vector<int>> d;
    queue<pair<TreeNode*, int>> q{{{root, 0}}};
    while (!q.empty()) {
      for (int n = q.size(); n; --n) {
        auto p = q.front();
        q.pop();
        root = p.first;
        int offset = p.second;
        d[offset].push_back(root->val);
        if (root->left) q.push({root->left, offset - 1});
        if (root->right) q.push({root->right, offset + 1});
      }
    }
    for (auto& [_, v] : d) {
      ans.push_back(v);
    }
    return ans;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func verticalOrder(root *TreeNode) [][]int {
  ans := [][]int{}
  if root == nil {
    return ans
  }
  d := map[int][]int{}
  q := []pair{pair{root, 0}}
  for len(q) > 0 {
    for n := len(q); n > 0; n-- {
      p := q[0]
      q = q[1:]
      root = p.node
      offset := p.offset
      d[offset] = append(d[offset], root.Val)
      if root.Left != nil {
        q = append(q, pair{root.Left, offset - 1})
      }
      if root.Right != nil {
        q = append(q, pair{root.Right, offset + 1})
      }
    }
  }
  idx := []int{}
  for i := range d {
    idx = append(idx, i)
  }
  sort.Ints(idx)
  for _, i := range idx {
    ans = append(ans, d[i])
  }
  return ans
}

type pair struct {
  node   *TreeNode
  offset int
}