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？

Posted 2021-04-23 福大大架构师每日一题

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：

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：

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[左神java代码](https://github.com/algorithmzuo/algorithmbasic2020/blob/master/src/class30/Code05_MinHeight.java#L35)

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