1. 题目
2. 思路
(1) DFS（递归）
- 若结点为null，则返回0，否则返回左子树和右子树中较大的高度值加1。
(2) BFS
3. 代码
import java.util.LinkedList;
import java.util.Queue;
public class Test {
public static void main(String[] args) {
}
}
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;
}
}
class Solution1 {
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}
int res = 0;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
res++;
}
return res;
}
}