二叉树深度
1: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。
例如:
给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回它的最大深度 3 。
提示:
节点总数 <= 10000
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof
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2:题目分析
改写前序遍历,统计每次下次向下递归深度+1,每次回溯的时候-1。到达叶子节点时统计深度比较即可。
3:代码示例
package JianZhiOffer55; /** * @author :dazhu * @date :Created in 2020/4/10 17:40 * @description: * @modified By: * @version: $ */ public class Main { public static void main(String[]args){ TreeNode n0 = new TreeNode(0); TreeNode n1 = new TreeNode(1); TreeNode n2 = new TreeNode(2); TreeNode n3 = new TreeNode(3); TreeNode n4 = new TreeNode(4); TreeNode n5 = new TreeNode(5); TreeNode n6 = new TreeNode(6); n0.right = n1; n0.left = n2; n1.right = n3; n1.left = n4; n3.right = n5; n5.left = n6; Solution solution = new Solution(); System.out.println(solution.maxDepth(n0)); } } class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } } class Solution { public int maxDepth(TreeNode root) { recur(root); return resultLength; } //改写常规的前序遍历方法 //每次向下递归时,深度+1。向上回溯时-1 //到达叶子节点时的记录深度即可。 private int length = 0; private int resultLength = 0; public void recur(TreeNode root) { if(root == null){ return; } //向下递归+1 length++; if(root.left==null&&root.right==null){ if(length>resultLength){ resultLength = length; } //到达叶子节点后,回溯之前,深度-1 length--; return; } recur(root.left); recur(root.right); //回溯-1 length--; } }