# go 二叉树的非递归遍历
## code TreeNode
```go
type TreeNode struct {
Val int
Left *TreeNode
Right *TreeNode
}
```
## 先序遍历 中左右
### code
```go
func preorderTraversal(root *TreeNode) []int {
if root == nil {
return []int{}
}
res := []int{}
stack := []*TreeNode{}
cur := root
for len(stack) > 0 || cur != nil {
if cur != nil {
res = append(res, cur.Val)
stack = append(stack, cur)
cur = cur.Left
} else {
cur = stack[len(stack)-1]
stack = stack[:len(stack)-1]
cur = cur.Right
}
}
return res
}
```
## 中序遍历 左中右
于先序遍历一样先将左边节点入栈
### code
```go
func inorderTraversal(root *TreeNode) []int {
if root == nil {
return []int{}
}
res := []int{}
stack := []*TreeNode{}
cur := root
for len(stack) > 0 || cur != nil {
for cur != nil {
stack = append(stack, cur)
cur = cur.Left
}
if len(stack) > 0 {
cur = stack[len(stack)-1]
res = append(res, cur.Val)
stack = stack[:len(stack)-1]
cur = cur.Right
}
}
return res
}
```
## 后续遍历 左右中
**注意**：“左右中”就是“中右左”的逆序，而按照“中右左” 顺序遍历二叉树的方思路和 先序“中左右”一样
### code
```go
func postorderTraversal(root *TreeNode) []int {
if root == nil {
return []int{}
}
res := []int{}
stack := []*TreeNode{}
cur := root
for len(stack) > 0 || cur != nil {
if cur != nil {
res = append(res, cur.Val)
stack = append(stack, cur)
cur = cur.Right
} else {
cur = stack[len(stack)-1]
stack = stack[:len(stack)-1]
cur = cur.Left
}
}
reverse(res)
return res
}
func reverse(res []int) {
for i, j := 0, len(res)-1; i < j; i, j = i+1, j-1 {
res[i], res[j] = res[j], res[i]
}
}
```
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