Let's make this one less mysterious. For 897. Increasing Order Search Tree, the solution in this repository is mainly a two pointers solution.
Guide
What?
We want to turn the problem statement into a smaller set of decisions the computer can repeat safely. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: two pointers, stack, sliding window.
Guide
When?
This pattern shows up when the brute force version has too many repeated checks, too many possible branches, or too much bookkeeping to do by hand. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are increasingBST, inOrder.
Guide
Why?
The point of the implementation is not to make the code longer. It is to avoid doing the same thinking twice.
- The stack stores unfinished context, which is usually the cleanest way to handle nested or monotonic structure.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Initialize the memory or helper structure.
- Process candidates in the order the invariant expects.
- Update the answer only when the current state is valid.
- Return the value that represents the fully processed input.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
solution.cpp
01/**
02Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.
03
04Example 1:
05Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]
06
07 5
08 / \
09 3 6
10 / \ \
11 2 4 8
12 / / \
131 7 9
14
15Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
16
17 1
18 \
19 2
20 \
21 3
22 \
23 4
24 \
25 5
26 \
27 6
28 \
29 7
30 \
31 8
32 \
33 9
34Note:
35
36The number of nodes in the given tree will be between 1 and 100.
37Each node will have a unique integer value from 0 to 1000.
38**/
39
40//Runtime: 72 ms, faster than 56.90% of C++ online submissions for Increasing Order Search Tree.
41//Memory Usage: 37.7 MB, less than 5.68% of C++ online submissions for Increasing Order Search Tree.
42/**
43 * Definition for a binary tree node.
44 * struct TreeNode {
45 * int val;
46 * TreeNode *left;
47 * TreeNode *right;
48 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
49 * };
50 */
51class Solution {
52public:
53 TreeNode* increasingBST(TreeNode* root) {
54 //in-order traversal
55 stack<TreeNode*> stk;
56 TreeNode* cur = root;
57 TreeNode* newcur = NULL;
58 TreeNode* newroot = NULL;
59 bool isFirst = true;
60
61 while(cur!=NULL || stk.size() > 0){
62 while(cur!=NULL){
63 stk.push(cur);
64 cur = cur->left;
65 }
66 cur = stk.top();
67 stk.pop();
68
69 TreeNode node = TreeNode(cur->val);
70 if(isFirst){
71 newroot = new TreeNode(cur->val);
72 newcur = newroot;
73 isFirst = !isFirst;
74 }else{
75 newcur->right = new TreeNode(cur->val);
76 newcur = newcur->right;
77 }
78
79 cur = cur->right;
80 }
81
82 return newroot;
83 }
84};
85
86//Approach 1: In-Order Traversal
87//Runtime: 72 ms, faster than 56.90% of C++ online submissions for Increasing Order Search Tree.
88//Memory Usage: 37 MB, less than 18.44% of C++ online submissions for Increasing Order Search Tree.
89/**
90Complexity Analysis
91
92Time Complexity: O(N)O(N), where NN is the number of nodes in the given tree.
93
94Space Complexity: O(N)O(N), the size of the answer.
95**/
96/**
97 * Definition for a binary tree node.
98 * struct TreeNode {
99 * int val;
100 * TreeNode *left;
101 * TreeNode *right;
102 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
103 * };
104 */
105
106 /**
107class Solution {
108public:
109 void inOrder(TreeNode* root, vector<int>& values){
110 if(root==NULL) return;
111 inOrder(root->left, values);
112 values.push_back(root->val);
113 inOrder(root->right, values);
114 }
115
116 TreeNode* increasingBST(TreeNode* root) {
117 vector<int> values;
118 inOrder(root, values);
119
120 TreeNode* newcur = new TreeNode(0);
121 TreeNode* newroot = newcur;
122
123 for(auto val : values){
124 newcur->right = new TreeNode(val);
125 newcur = newcur->right;
126 }
127
128 return newroot->right;
129 }
130};
131**/
132
133/**
134Approach 2: Traversal with Relinking
135Intuition and Algorithm
136
137We can perform the same in-order traversal as in Approach 1.
138During the traversal, we'll construct the answer on the fly,
139reusing the nodes of the given tree by cutting their left child and adjoining them to the answer.
140**/
141
142/**
143Time Limit Exceeded
144Last executed input
145[379
146826]
147**/
148
149/**
150class Solution {
151public:
152 TreeNode* newcur;
153
154 void inOrder(TreeNode* node){
155 if(node==NULL) return;
156 inOrder(node->left);
157 newcur->left = NULL;
158 newcur->right = node;
159 newcur = node;
160 inOrder(node->right);
161 }
162
163 TreeNode* increasingBST(TreeNode* root) {
164 TreeNode* newroot = new TreeNode(0);
165 newcur = newroot;
166 inOrder(root);
167 return newroot->right;
168 }
169};
170**/Cost
Complexity
Time
O(n) to O(n log n), depending on the dominant loop or data structure operation
Dominated by the main traversal, recursion, or data-structure operations in the code.
Space
O(n) in the usual case for auxiliary containers or recursion
Auxiliary state plus the answer structure where the problem requires one.