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 mainimport "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：
左神java代码
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