111. 二叉树的最小深度

输入：root = [3,9,20,null,null,15,7]输出：2

输入：root = [2,null,3,null,4,null,5,null,6]输出：5

/** * Definition for a binary tree node. * type TreeNode struct { *     Val int *     Left *TreeNode *     Right *TreeNode * } */func minDepth(root *TreeNode) int {
	var queue []*TreeNode	var minDepth func(*TreeNode) int
	push := func(node *TreeNode) {		queue = append(queue, node)	}
	pop := func() *TreeNode {		tmp := queue[0]		queue = queue[1:]		return tmp	}
	minDepth = func(root *TreeNode) int {		if root == nil {			return 0		}
		queue = append(queue, root)		depth := 1		for len(queue) > 0 {			sz := len(queue)			for i := 0; i < sz; i++ {				node := pop()				if node.Left == nil && node.Right == nil { // 当左节点和右节点都是null的时候返回					return depth				}				if node.Left != nil {					push(node.Left)				}				if node.Right != nil {					push(node.Right)				}			}			depth++ // 每层结束会+1		}
		return depth	}   return minDepth(root)}

/** * Definition for a binary tree node. * type TreeNode struct { *     Val int *     Left *TreeNode *     Right *TreeNode * } */func minDepth(root *TreeNode) int {	res := 0	if root == nil {		return res	}	searchPath(root, 0, &res)	return res}
func searchPath(root *TreeNode, dep int, res *int) {	if root != nil {		dep++		if root.Left == nil && root.Right == nil {			if dep < *res || *res == 0 {				*res = dep			}		} else {			searchPath(root.Left, dep, res)			searchPath(root.Right, dep, res)		}	}}