跳转至

589. N 叉树的前序遍历

题目描述

给定一个 n 叉树的根节点  root ，返回 其节点值的 前序遍历 。

n 叉树 在输入中按层序遍历进行序列化表示，每组子节点由空值 null 分隔（请参见示例）。


示例 1：

输入：root = [1,null,3,2,4,null,5,6]
输出：[1,3,5,6,2,4]

示例 2：

输入：root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
输出：[1,2,3,6,7,11,14,4,8,12,5,9,13,10]

 

提示：

  • 节点总数在范围 [0, 104]内
  • 0 <= Node.val <= 104
  • n 叉树的高度小于或等于 1000

 

进阶：递归法很简单，你可以使用迭代法完成此题吗?

解法

方法一：递归

我们可以递归地遍历整棵树。对于每个节点，先将节点的值加入答案，然后对该节点的每个子节点递归地调用函数。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为节点数。

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""


class Solution:
    def preorder(self, root: "Node") -> List[int]:
        def dfs(root):
            if root is None:
                return
            ans.append(root.val)
            for child in root.children:
                dfs(child)

        ans = []
        dfs(root)
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    private List<Integer> ans = new ArrayList<>();

    public List<Integer> preorder(Node root) {
        dfs(root);
        return ans;
    }

    private void dfs(Node root) {
        if (root == null) {
            return;
        }
        ans.add(root.val);
        for (Node child : root.children) {
            dfs(child);
        }
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<int> preorder(Node* root) {
        vector<int> ans;
        function<void(Node*)> dfs = [&](Node* root) {
            if (!root) {
                return;
            }
            ans.push_back(root->val);
            for (auto& child : root->children) {
                dfs(child);
            }
        };
        dfs(root);
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func preorder(root *Node) (ans []int) {
    var dfs func(*Node)
    dfs = func(root *Node) {
        if root == nil {
            return
        }
        ans = append(ans, root.Val)
        for _, child := range root.Children {
            dfs(child)
        }
    }
    dfs(root)
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
/**
 * Definition for node.
 * class Node {
 *     val: number
 *     children: Node[]
 *     constructor(val?: number) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.children = []
 *     }
 * }
 */

function preorder(root: Node | null): number[] {
    const ans: number[] = [];
    const dfs = (root: Node | null) => {
        if (!root) {
            return;
        }
        ans.push(root.val);
        for (const child of root.children) {
            dfs(child);
        }
    };
    dfs(root);
    return ans;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
/**
 * Definition for a Node.
 * struct Node {
 *     int val;
 *     int numChildren;
 *     struct Node** children;
 * };
 */

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */

void dfs(struct Node* root, int* ans, int* i) {
    if (!root) {
        return;
    }
    ans[(*i)++] = root->val;
    for (int j = 0; j < root->numChildren; j++) {
        dfs(root->children[j], ans, i);
    }
}

int* preorder(struct Node* root, int* returnSize) {
    int* ans = malloc(sizeof(int) * 10000);
    *returnSize = 0;
    dfs(root, ans, returnSize);
    return ans;
}

方法二：迭代（栈实现）

我们也可以用迭代的方法来解决这个问题。

我们使用一个栈来帮助我们得到前序遍历，我们首先把根节点入栈，因为前序遍历是根节点、左子树、右子树，栈的特点是先进后出，所以我们先把节点的值加入答案，然后对该节点的每个子节点按照从右到左的顺序依次入栈。循环直到栈为空。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为节点数。

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, children=None):
        self.val = val
        self.children = children
"""


class Solution:
    def preorder(self, root: 'Node') -> List[int]:
        ans = []
        if root is None:
            return ans
        stk = [root]
        while stk:
            node = stk.pop()
            ans.append(node.val)
            for child in node.children[::-1]:
                stk.append(child)
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public List<Integer> preorder(Node root) {
        if (root == null) {
            return Collections.emptyList();
        }
        List<Integer> ans = new ArrayList<>();
        Deque<Node> stk = new ArrayDeque<>();
        stk.push(root);
        while (!stk.isEmpty()) {
            Node node = stk.pop();
            ans.add(node.val);
            List<Node> children = node.children;
            for (int i = children.size() - 1; i >= 0; --i) {
                stk.push(children.get(i));
            }
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<int> preorder(Node* root) {
        if (!root) return {};
        vector<int> ans;
        stack<Node*> stk;
        stk.push(root);
        while (!stk.empty()) {
            Node* node = stk.top();
            ans.push_back(node->val);
            stk.pop();
            auto children = node->children;
            for (int i = children.size() - 1; i >= 0; --i) stk.push(children[i]);
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func preorder(root *Node) (ans []int) {
    if root == nil {
        return
    }
    stk := []*Node{root}
    for len(stk) > 0 {
        node := stk[len(stk)-1]
        ans = append(ans, node.Val)
        stk = stk[:len(stk)-1]
        children := node.Children
        for i := len(children) - 1; i >= 0; i-- {
            stk = append(stk, children[i])
        }
    }
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
/**
 * Definition for node.
 * class Node {
 *     val: number
 *     children: Node[]
 *     constructor(val?: number) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.children = []
 *     }
 * }
 */

function preorder(root: Node | null): number[] {
    const ans: number[] = [];
    if (!root) {
        return ans;
    }
    const stk: Node[] = [root];
    while (stk.length) {
        const { val, children } = stk.pop()!;
        ans.push(val);
        for (let i = children.length - 1; i >= 0; i--) {
            stk.push(children[i]);
        }
    }
    return ans;
}

评论