2021-03-21：给定一棵二叉树的头节点head，求以head为头的树中，最小深度是多少？
福大大 答案2021-03-21：
1.递归。
2.莫里斯遍历。
代码用golang编写，代码如下：
代码语言：txt复制package main
import "fmt"
func main() {
head := &TreeNode{}
head.Left = &TreeNode{}
head.Right = &TreeNode{}
head.Right.Right = &TreeNode{}
ret := minHeight1(head)
fmt.Println("1.递归:", ret)
ret = minHeight2(head)
fmt.Println("2.莫里斯遍历：", ret)
}
//Definition for a binary tree node.
type TreeNode struct {
Val int
Left *TreeNode
Right *TreeNode
}
const INT_MAX = int(^uint(0) >> 1)
func minHeight1(head *TreeNode) int {
if head == nil {
return 0
}
leftVal := minHeight1(head.Left)
rightVal := minHeight1(head.Right)
return 1 getMin(leftVal, rightVal)
}
// 根据morris遍历改写
func minHeight2(head *TreeNode) int {
if head == nil {
return 0
}
cur := head
var mostRight *TreeNode
curLevel := 0
minHeight := INT_MAX
for cur != nil {
mostRight = cur.Left
if mostRight != nil {
rightBoardSize := 1
for mostRight.Right != nil && mostRight.Right != cur {
rightBoardSize
mostRight = mostRight.Right
}
if mostRight.Right == nil { // 第一次到达
curLevel
mostRight.Right = cur
cur = cur.Left
continue
} else { // 第二次到达
if mostRight.Left == nil {
minHeight = getMin(minHeight, curLevel)
}
curLevel -= rightBoardSize
mostRight.Right = nil
}
} else { // 只有一次到达
curLevel
}
cur = cur.Right
}
finalRight := 1
cur = head
for cur.Right != nil {
finalRight
cur = cur.Right
}
if cur.Left == nil && cur.Right == nil {
minHeight = getMin(minHeight, finalRight)
}
return minHeight
}
func getMin(a int, b int) int {
if a < b {
return a
} else {
return b
}
}执行结果如下：
在这里插入图片描述左神java代码
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