leetcode-144-Binary-Tree-Preorder-Traversal

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描述

Given a binary tree, return the preorder traversal of its nodes’ values.

Example:

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Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,2,3]

Follow up: Recursive solution is trivial, could you do it iteratively?

分析

先序遍历，可以当做例题。

解决方案1（Python）

递归版本。

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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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        
        def preOrder(root):
            if root:
                result.append(root.val)
                preOrder(root.left)
                preOrder(root.right)
        preOrder(root)
        return result

解决方案2(Java)

递归版本。

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        preOrder(root, result);
        return result;
    }
    private void preOrder(TreeNode root, ArrayList<Integer> result) {
        if (root == null) {
            return;
        }
        result.add(root.val);
        preOrder(root.left, result);
        preOrder(root.right, result);
    }
    
}

解决方案3（golang）

递归版本。

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func preorderTraversal(root *TreeNode) []int {
    if root == nil {
        return []int{}
    }
    result := []int{}
    result = append(result, root.Val)
    left := append(result, preorderTraversal(root.Left)...)
    return append(left, preorderTraversal(root.Right)...)
}

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