Discuss：www.cnblogs.com/grandyang/p…
Given the root of a binary tree, return its maximum depth.
A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: 3
Example 2:
Input: root = [1,null,2]
Output: 2
Example 3:
Input: root = []
Output: 0
Example 4:
Input: root = [0]
Output: 1
Constraints:
- The number of nodes in the tree is in the range
[0, 104]. -100 <= Node.val <= 100
解法一：
DFS，深度优先搜索。求二叉树的最大深度问题用到深度优先搜索 Depth First Search，递归的完美应用。
class Solution {
fun maxDepth(root: TreeNode?): Int {
if (root == null) {
return 0
}
return 1 + maxDepth(root.left).coerceAtLeast(maxDepth(root.right))
}
}
解法二：
BFS，我们也可以使用层序遍历二叉树，然后计数总层数，即为二叉树的最大深度，参见代码如下：
class Solution {
fun maxDepth(root: TreeNode?): Int {
if (root == null) {
return 0
}
var result = 0
val queue = ArrayDeque<TreeNode>()
queue.offer(root)
while (queue.isNotEmpty()) {
val length = queue.size
result++
for (index in 0 until length) {
val currentNode = queue.poll()
currentNode.left?.let {
queue.offer(it)
}
currentNode.right?.let {
queue.offer(it)
}
}
}
return result
}
}
