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:
[0, 10^4]
?区间内。-100 <= Node.val <= 100
1、递归/深度优先遍历(Recursion,Depth-First Traversal)
2、迭代/广度优先遍历(Iterator,Breadth-First Traversal)
法一:
/**
* 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) {
// Recursion,Depth-First Traversal
// Time: O(n) n 节点数
// Space: O(h) h 高度
if (root == null) return 0;
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}
}
法二:
/**
* 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) {
// Iterator, Breadth-First Traversal
// Time: O(n) n 节点数
// Space: O(n) 每一层最大的节点数,最好n,最坏1
if (root == null) {
return 0;
}
Deque<TreeNode> queue = new ArrayDeque<>();
queue.addLast(root);
int depth = 0;
while (!queue.isEmpty()) {
int size = queue.size();
while (size-- > 0) {
TreeNode node = queue.removeFirst();
if (node.left != null) {
queue.addLast(node.left);
}
if (node.right != null) {
queue.addLast(node.right);
}
}
depth++;
}
return depth;
}
}