题目描述:
输入一棵二叉树的根节点,求该树的深度。从根节点到叶节点依次经过的节点(含根、叶节点)形成树的一条路径,最长路径的长度为树的深度。
题目来源于LeeCode:https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof/
解题思路:
- 方法一:后序遍历:递归的找到左右子树的深度,然后返回最大值。注意:别忘记最后要加上根节点的深度为1。
- 方法二:层序遍历:二叉树的深度等于二叉树的层数,所以层序遍历二叉树,统计二叉树的层数即可。
后序遍历:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) return 0;
return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}
}
层序遍历:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) return 0;
Queue<TreeNode> arr = new LinkedList<TreeNode>();
arr.offer(root);
int depth = 0;
while (!arr.isEmpty()) {
Queue<TreeNode> ans = new LinkedList<TreeNode>();
int len = arr.size();
for (int i = 0; i < len; i++) {
TreeNode cur = arr.poll();
if (cur.left != null) {
ans.offer(cur.left);
}
if (cur.right != null) {
ans.offer(cur.right);
}
}
arr = ans;
depth++;
}
return depth;
}
}