Tree - 145. Binary Tree Postorder Traversal

145. Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
1

2
/
3
Output: [3,2,1]

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

思路：

二叉树后序遍历，递归和迭代来做，迭代也是使用list容器来充当栈，后序遍历和前序后序遍历主要特点在于父节点会被访问两次，所以使用一个辅助指针preVisited来记录是否是第二次访问这个节点。

代码：

go：

/**

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

/* // recursive
func postorderTraversal(root *TreeNode) []int {
    var res []int
    
    recursive(root, &res)
    
    return res
}


func recursive(root *TreeNode, res *[]int) {
    if root == nil {
        return
    }
    
    recursive(root.Left, res)
    recursive(root.Right, res)
    *res = append(*res, root.Val)
}
*/

// iteratively
func postorderTraversal(node *TreeNode) []int {

    var res []int 
    var stack []*TreeNode
    var visited *TreeNode
    
    for node != nil || len(stack) != 0 {
        for node != nil {
            stack = append(stack, node)
            node = node.Left
        }
        
        // 访问栈顶元素
        node = stack[len(stack) - 1]
        
        if node.Right != nil && node.Right != visited {
            node = node.Right
        } else {
            // 父节点的右子树被访问过了，父节点出栈
            stack = stack[:len(stack) - 1]
            res = append(res, node.Val)
            visited = node
            node = nil  // 保证不会再被入栈
        }
    }
    
    return res
}