Question

589. N 叉树的前序遍历

English Version

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。

"""
# 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

/*
// 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);
    }
  }
}

/*
// 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

/**
 * 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。

"""
# 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

/*
// 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;
  }
}

/*
// 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;
  }
};

/**
 * 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
}

/**
 * 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;
}