0144. Binary Tree Preorder Traversal

144. Binary Tree Preorder Traversal #

题目 #

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

Example:


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

Output: [1,2,3]

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

题目大意 #

先根遍历一颗树。

解题思路 #

两种递归的实现方法,见代码。

代码 #


package leetcode

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

// 解法一 递归
func preorderTraversal(root *TreeNode) []int {
	res := []int{}
	if root != nil {
		res = append(res, root.Val)
		tmp := preorderTraversal(root.Left)
		for _, t := range tmp {
			res = append(res, t)
		}
		tmp = preorderTraversal(root.Right)
		for _, t := range tmp {
			res = append(res, t)
		}
	}
	return res
}

// 解法二 递归
func preorderTraversal1(root *TreeNode) []int {
	var result []int
	preorder(root, &result)
	return result
}

func preorder(root *TreeNode, output *[]int) {
	if root != nil {
		*output = append(*output, root.Val)
		preorder(root.Left, output)
		preorder(root.Right, output)
	}
}

// 解法三 非递归,用栈模拟递归过程
func preorderTraversal2(root *TreeNode) []int {
	if root == nil {
		return []int{}
	}
	stack, res := []*TreeNode{}, []int{}
	stack = append(stack, root)
	for len(stack) != 0 {
		node := stack[len(stack)-1]
		stack = stack[:len(stack)-1]
		if node != nil {
			res = append(res, node.Val)
		}
		if node.Right != nil {
			stack = append(stack, node.Right)
		}
		if node.Left != nil {
			stack = append(stack, node.Left)
		}
	}
	return res
}


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