🥈Medium

给定一个二叉树，返回它的 前序 遍历。
示例:

    输入: [1,null,2,3]
  1
    \
     2
    /
   3
输出: [1,2,3]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

题解

递归

递归还是挺好写的，只要记住先序顺序为root->left->right即可

Python

# 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 preorderTraversal(self, root: TreeNode) -> List[int]:
        def preorder(node):
            if not node:
                return 
            res.append(node.val)
            preorder(node.left)
            preorder(node.right)
        res = []
        preorder(root)
        return res

一行解决

class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        return [root.val] + self.preorderTraversal(root.left) + self.preorderTraversal(root.right)

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var preorderTraversal = function(root) {
    const res = []
    const preOrder = (root) => {
        if (root == null) return
        res.push(root.val)
        preOrder(root.left)
        preOrder(root.right)
    };
    preOrder(root)
    return res
};

复杂度分析

• 时间复杂度： 其中n是二叉树节点数，每个节点恰好被遍历一次
• 空间复杂度： 为递归过程中的开销。平均情况下为，最坏情况下树为链状，为

  迭代

  迭代过程如图所示
  
  
  
  
  
  
  
  
  
  
  
  
  
  

  Python

  ```python

  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 preorderTraversal(self, root: TreeNode) -> List[int]:

    res=[]
    if not root:
        return res
    stack = [root]
    while(stack):
        cur=stack.pop()
        res.append(cur.val)
        if cur.right:
            stack.append(cur.right)
        if cur.left:
            stack.append(cur.left)
    return res
  

<a name="l4jHb"></a>
#### JavaScript
```javascript
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
const preorderTraversal = (root) => {
  if (root === null){
      return []
  }
  let ans=[]
  let stack=[root]
  while(stack.length>0){
      let node = stack.pop()
      ans.push(node.val)
      if(node.right){
          stack.push(node.right)
      }
      if(node.left){
          stack.push(node.left)
      }

  }
  return ans
}

复杂度分析

• 时间复杂度： 其中n是二叉树节点数，每个节点恰好被遍历一次
• 空间复杂度： 为迭代过程中显式栈的开销。平均情况下为，最坏情况下树为链状，为