题目
给定一个二叉树 root ,返回其最大深度。
二叉树的 最大深度 是指从根节点到最远叶子节点的最长路径上的节点数。
示例 1:

输入:root = [3,9,20,null,null,15,7]
输出:3
示例 2:
输入:root = [1,null,2]
输出:2
提示:
- 树中节点的数量在
[0, 104]区间内。 -100 <= Node.val <= 100
题解
题解一(DFS后序遍历)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if(root == null) return 0;
return Math.max(maxDepth(root.left),maxDepth(root.right)) + 1;
}
}
题解二(BFS层序遍历)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) return 0;
List<TreeNode> queue = new LinkedList<>();
queue.add(root);
List<TreeNode> tmp = new LinkedList<>();
int res = 0;
while (!queue.isEmpty()) {
tmp = new LinkedList<>();
for(TreeNode node : queue) {
if (node.left != null) tmp.add(node.left);
if (node.right != null) tmp.add(node.right);
}
queue = tmp;
res++;
}
return res;
}
}