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314. 二叉树的垂直遍历 🔒

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 = [1,2,3,4,10,9,11,null,5,null,null,null,null,null,null,null,6]
输出：[[4],[2,5],[1,10,9,6],[3],[11]]

 

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。

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# 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

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/**
 * 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);
    }
}

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/**
 * 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);
    }
};

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/**
 * 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。

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# 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())]

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/**
 * 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());
    }
}

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/**
 * 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;
    }
};

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/**
 * 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
}

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