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111. 二叉树的最小深度 #92

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@Geekhyt

Description

原题链接

递归 dfs

  1. root 为空时,高度为 0
  2. root 的左右子树都为空时,高度为 1
  3. 如果左子树或者右子树为空时,返回另一棵子树的高度
  4. 否则返回两棵子树的高度最小值
const minDepth = function(root) {
 if (root === null) return 0
 if (root.left === null && root.right === null) {
 return 1
 }
 if (root.left === null) {
 return 1 + minDepth(root.right)
 }
 if (root.right === null) {
 return 1 + minDepth(root.left)
 }
 return Math.min(minDepth(root.left), minDepth(root.right)) + 1
}
  • 时间复杂度:O(n)
  • 空间复杂度:O(logn)

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