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#yyds干货盘点# 面试必刷TOP101：二叉树的中序遍历
source link: https://blog.51cto.com/u_15488507/5587685
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#yyds干货盘点# 面试必刷TOP101：二叉树的中序遍历
精选 原创1.简述：
描述给定一个二叉树的根节点root，返回它的中序遍历结果。
数据范围：树上节点数满足 ，树上每个节点的值满足 进阶：空间复杂度 ，时间复杂度
示例1{1,2,#,#,3}
[2,3,1]
{1,2}
[2,1]
{1,#,2}
[1,2]
2.代码实现：
import java.util.*;
public class Solution {
public void inorder(List<Integer> list, TreeNode root){
//遇到空节点则返回
if(root == null)
return;
//先去左子树
inorder(list, root.left);
//再访问根节点
list.add(root.val);
//最后去右子树
inorder(list, root.right);
}
public int[] inorderTraversal (TreeNode root) {
//添加遍历结果的数组
List<Integer> list = new ArrayList();
//递归中序遍历
inorder(list, root);
//返回的结果
int[] res = new int[list.size()];
for(int i = 0; i < list.size(); i++)
res[i] = list.get(i);
return res;
}
}
public class Solution {
public void inorder(List<Integer> list, TreeNode root){
//遇到空节点则返回
if(root == null)
return;
//先去左子树
inorder(list, root.left);
//再访问根节点
list.add(root.val);
//最后去右子树
inorder(list, root.right);
}
public int[] inorderTraversal (TreeNode root) {
//添加遍历结果的数组
List<Integer> list = new ArrayList();
//递归中序遍历
inorder(list, root);
//返回的结果
int[] res = new int[list.size()];
for(int i = 0; i < list.size(); i++)
res[i] = list.get(i);
return res;
}
}
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