[Golang][Leetcode][BinaryTree]刷題系列-107-Level Order Traversal II

2021-08-25 292 words 2 minutes

/leetcode107_binarytree/golang_leetcode.png

Contents

107. Binary Tree Level Order Traversal II

Level : Medium

原題連結 : Click

題目 :

Given the root of a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from leaf to root).

Example :

Note

Example 1:

Input: root = [3,9,20,null,null,15,7] Output: [[15,7],[9,20],[3]]

Example 2:

Input: root = [1] Output: [[1]]

Example 3:

Input: root = [] Output: []

解題思路 :

這題可以直接參考 leetcode102題解，使用到的觀念一模一樣，只是多一個反轉結果而已!!!

Queue解法

time complexity: O(n) , space complexity: O(n)

Runtime: 0 ms, faster than 100.00% of Go online submissions for Binary Tree Level Order Traversal II.

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrderBottom(root *TreeNode) [][]int {
    
    var res [][]int
    
    if root == nil {
        return res
    }
    
    var queue []*TreeNode
    
    queue = append(queue, root)
    
    for len(queue) != 0 {
        
        length := len(queue)
        
        tempArr := make([]int,0)
        
        for i := 0; i < length; i++ {
            
            node := queue[0]
            
            if node.Left != nil {
                queue = append(queue, node.Left)
            }
            
            if node.Right != nil {
                queue = append(queue, node.Right)
            }
            
            tempArr = append(tempArr, node.Val)
            
            queue = queue[1:]
            
        }
        
        res = append(res, tempArr)
    }
    
    for i, j := 0, len(res)-1; i < j; i, j = i+1, j-1 {
        res[i], res[j] = res[j], res[i]
    }
    
    return res
}

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