原題鏈接：https://leetcode.com/problems/binary-tree-postorder-traversal/
難度：hard
解法：二叉樹的后序遍歷，左孩子 -》右孩子-》父節(jié)點(diǎn)。挺簡單的，不知道為啥算是hard難度。
編程語言：golang

遞歸版實(shí)現(xiàn)：

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

    if root.Left != nil {
        r := postorderTraversal(root.Left)
        for _, ele := range r {
            result = append(result, ele)
        }
    }
    
    if root.Right != nil {
        r := postorderTraversal(root.Right)
        for _, ele := range r {
            result = append(result, ele)
        }
    }

    return append(result, root.Val)
}

非遞歸版實(shí)現(xiàn)：利用兩個(gè)棧來完成后續(xù)遍歷。go沒有內(nèi)建的stack也是挺煩的，pop操作，一行代碼變兩行了。

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func postorderTraversal(root *TreeNode) []int {
    var result []int
    
    if root == nil {
        return result
    }
    
    var s1 []*TreeNode
    var s2 []*TreeNode
    
    s1 = append(s1, root);
    
    for len(s1) > 0 {
        // use slick to do stack pop operation
        n := s1[len(s1) - 1]
        s1 = s1[0:len(s1) - 1]
        
        if n.Left != nil {
            s1 = append(s1, n.Left);
        }
        if n.Right != nil {
            s1 = append(s1, n.Right);
        }
        s2 = append(s2, n);
    }
    
    for len(s2) > 0 {
        // pop the header of s2
        n := s2[len(s2) - 1]
        s2 = s2[0:len(s2) - 1]
        result = append(result, n.Val)
    }

    return result
}

非遞歸版實(shí)現(xiàn)，看起來時(shí)間和內(nèi)存占用都還不錯(cuò)：
Runtime: 0 ms, faster than 100.00% of Go online submissions for Binary Tree Postorder Traversal.
Memory Usage: 2 MB, less than 100.00% of Go online submissions for Binary Tree Postorder Traversal.