-
Notifications
You must be signed in to change notification settings - Fork 44
/
_94_inorderTraversal.java
95 lines (87 loc) · 2.42 KB
/
_94_inorderTraversal.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
package pp.arithmetic.leetcode;
import pp.arithmetic.Util;
import pp.arithmetic.model.TreeNode;
import java.util.ArrayList;
import java.util.List;
import java.util.Stack;
/**
* Created by wangpeng on 2019-07-24.
* 94. 二叉树的中序遍历
* <p>
* 给定一个二叉树,返回它的中序 遍历。
* <p>
* 示例:
* <p>
* 输入: [1,null,2,3]
* 1
* \
* 2
* /
* 3
* <p>
* 输出: [1,3,2]
* 进阶: 递归算法很简单,你可以通过迭代算法完成吗?
* <p>
* <p>
* 来源:力扣(LeetCode)
* 链接:https://leetcode-cn.com/problems/binary-tree-inorder-traversal
* 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
*/
public class _94_inorderTraversal {
public static void main(String[] args) {
TreeNode treeNode = Util.generateTreeNode();
Util.printTree(treeNode);
_94_inorderTraversal inorderTraversal = new _94_inorderTraversal();
//递归算法
Util.printList(inorderTraversal.inorderTraversal(treeNode));
Util.printList(inorderTraversal.inorderTraversal2(treeNode));
}
/**
* 解题思路:
* 按题目所说,递归算法{@link _94_inorderTraversal#inorderTraversal2(TreeNode)}很简单,我们试着迭代算法完成
* 模拟递归,用一个栈保存遍历结果
*
* @param root
* @return
*/
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
//左
while (curr != null) {
stack.push(curr);
curr = curr.left;
}
curr = stack.pop();
//中
result.add(curr.val);
//右
curr = curr.right;
}
return result;
}
/**
* 递归算法
*
* @param root
* @return
*/
public List<Integer> inorderTraversal2(TreeNode root) {
List<Integer> result = new ArrayList<>();
recursion(root, result);
return result;
}
private void recursion(TreeNode root, List<Integer> result) {
if (root == null) {
return;
}
//左
recursion(root.left, result);
//中
result.add(root.val);
//右
recursion(root.right, result);
}
}