606. 根据二叉树创建字符串

606. 根据二叉树创建字符串

1 先序遍历采用栈结构，每次把根节点出栈的时候 按照先右节点再左节点的方式入栈。
2 之前的先序遍历只是把节点数据入栈即可，这次我们需要涉及到括号 ，所以需要每次入栈的时候增加括号，按照先右括号 在左括号的方式入栈
3 唯一需要特殊处理的是 当左节点为空 但是右节点不为空的情况时候 需要入栈一个没有值的空括号

package main
import (
    "fmt"
    "strings"
)
type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}
func tree2str(t *TreeNode) string {
    var str strings.Builder
    if t == nil {
        return ""
    }
    stack := []interface{}{t}
    for len(stack) > 0 {
        node := stack[len(stack)-1]
        stack = stack[:len(stack)-1]
        if n, ok := node.(*TreeNode); ok {
            str.WriteString(fmt.Sprintf("%d", n.Val))
            if n.Right != nil {
                stack = append(stack, ")")
                stack = append(stack, n.Right)
                stack = append(stack, "(")
            }
            if n.Right != nil && n.Left == nil {
                stack = append(stack, "()")
            }
            if n.Left != nil {
                stack = append(stack, ")")
                stack = append(stack, n.Left)
                stack = append(stack, "(")
            }
        } else {
            s := node.(string)
            str.WriteString(s)
        }
    }
    return str.String()
}
func main() {
    one := &TreeNode{Val: 1}
    two := &TreeNode{Val: 2}
    one.Left = two
    three := &TreeNode{Val: 3}
    one.Right = three
    four := &TreeNode{Val: 4}
    two.Right = four
    fmt.Println(tree2str(one))
    //one1 :=&TreeNode{Val: 1}
    //two1 :=&TreeNode{Val: 2}
    //one1.Left = two1
    //three1 :=&TreeNode{Val: 3}
    //one1.Right = three1
    //four1 :=&TreeNode{Val: 4}
    //two1.Right = four1
    //fmt.Println(tree2str(one1))
}