-
Notifications
You must be signed in to change notification settings - Fork 1
/
LeetCode-94-Binary-Tree-Inorder-Traversal.java
166 lines (141 loc) · 5.41 KB
/
LeetCode-94-Binary-Tree-Inorder-Traversal.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
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
/*
LeetCode: https://leetcode.com/problems/binary-tree-inorder-traversal/
LintCode: http://www.lintcode.com/problem/binary-tree-inorder-traversal/
JiuZhang: http://www.jiuzhang.com/solutions/binary-tree-inorder-traversal/
ProgramCreek: http://www.programcreek.com/2012/12/leetcode-solution-of-binary-tree-inorder-traversal-in-java/
Analysis:
1.DFS
2.BFS()
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
// 1.DFS
// public List<Integer> inorderTraversal(TreeNode root) {
// List<Integer> result = new ArrayList<Integer>();
// inorderTraversal(root, result);
// return result;
// }
// private void inorderTraversal(TreeNode root, List<Integer> result){
// if(root == null) return;
// if(root.left != null) inorderTraversal(root.left, result);
// result.add(root.val);
// if(root.right != null) inorderTraversal(root.right, result);
// }
// Another way to write without helper function
// public List<Integer> inorderTraversal(TreeNode root) {
// List<Integer> result = new ArrayList<>();
// if (root == null) return result;
//
// if (root.left != null) result.addAll(inorderTraversal(root.left));
// result.add(root.val);
// if (root.right != null) result.addAll(inorderTraversal(root.right));
//
// return result;
// }
// 2.DFS with stack
// public List<Integer> inorderTraversal(TreeNode root) {
// List<Integer> result = new ArrayList<Integer>();
// if(root == null) return result;
// Stack<TreeNode> stack = new Stack<TreeNode>();
// TreeNode curr = root;
// while(curr != null || !stack.isEmpty()){
// // push curr node's left tree into stack
// while(curr != null){
// stack.push(curr);
// curr = curr.left;
// }
// curr = stack.pop();
// result.add(curr.val);
// curr = curr.right;
// }
// return result;
// }
// 3.DFS with Stack(easier to understand than 2.)
// public List<Integer> inorderTraversal(TreeNode root) {
// List<Integer> result = new ArrayList<Integer>();
// if(root == null)return result;
//
// Stack<TreeNode> stack = new Stack<TreeNode>();
// TreeNode curr = root; //define a pointer to track nodes
//
// while(curr != null || !stack.isEmpty()){
// // if curr node is not null, push to stack, and go down the tree to left
// if(curr != null){
// stack.push(curr);
// curr = curr.left;
// }else{
// // if no left node, pop last left node, process it
// // then let curr point to right node
// curr = stack.pop();
// result.add(curr.val);
// curr = curr.right;
// }
// }
//
// return result;
// }
// DFS, trying to do it without additional pointer (This approach will add null to the stack.)
// public List<Integer> inorderTraversal(TreeNode root) {
// List<Integer> result = new ArrayList<>();
// if (root == null) return result;
// Stack<TreeNode> stack = new Stack<>();
// stack.push(root);
// while (!stack.isEmpty()) {
// TreeNode curr = stack.peek();
// if (curr != null) {
// stack.push(curr.left);
// } else {
// // Means no left node, pop last left node, and process it
// // then left curr point to right node
// stack.pop(); // pop last null element
// if (!stack.isEmpty()) {
// curr = stack.pop();
// // process curr
// result.add(curr.val);
// // After we proceed the last left element, let's move right
// stack.push(curr.right);
// }
// }
// }
// return result;
// }
// 4. Morris Inorder Traversal
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();
if (root == null) return result;
TreeNode curr = root;
while (curr != null) {
if (curr.left == null) {
// No left substree, so process curr, and move to right subtree
result.add(curr.val);
curr = curr.right;
} else {
// find the predecessor of curr node (the most right in left tree)
TreeNode pre = curr.left;
while (pre.right != null && pre.right != curr) {
pre = pre.right;
}
if (pre.right == null) {
pre.right = curr;
curr = curr.left;
} else {
// finished traverse all the left tree, so process root now
// reset the predecessor's right to null
// as we finished traverse all left tree, and root, we should move to right tree
result.add(curr.val);
pre.right = null;
curr = curr.right;
}
}
}
return result;
}
}