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btree_result.txt
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run btree-test.out
Test 1: is tree a binary search tree
The tree topology:
4 (L=3, R=3)
7 (L=1, R=2)
8 (L=0, R=1)
9 (L=0, R=0)
-
-
-
6 (L=0, R=0)
-
-
2 (L=2, R=1)
3 (L=0, R=0)
-
-
-1 (L=1, R=1)
0 (L=0, R=0)
-
-
-2 (L=0, R=0)
-
-
After deleting 2 and 7, the tree topology:
4 (L=3, R=2)
8 (L=1, R=1)
9 (L=0, R=0)
-
-
6 (L=0, R=0)
-
-
0 (L=2, R=1)
3 (L=0, R=0)
-
-
-1 (L=1, R=0)
-
-2 (L=0, R=0)
-
-
It is still a binary search tree
Test 2: traverse tree in level left to right
---- level = 0, pos = 0, val = 4
---- level = 1, pos = 1, val = 0
---- level = 1, pos = 2, val = 8
---- level = 2, pos = 3, val = -1
---- level = 2, pos = 4, val = 3
---- level = 2, pos = 5, val = 6
---- level = 2, pos = 6, val = 9
---- level = 3, pos = 7, val = -2
Test 3: find the next node of others in same level
The next node of 0 is 8
The next node of 6 is 9
The next node of -2 is none
Test 4: print values in reversed order of level
-2 -1 3 6 9 0 8 4
The tree topology:
50 (L=2, R=3)
60 (L=1, R=2)
70 (L=1, R=1)
80 (L=0, R=0)
-
-
65 (L=0, R=0)
-
-
45 (L=0, R=0)
-
-
30 (L=1, R=1)
20 (L=0, R=0)
-
-
5 (L=0, R=0)
-
-
Test 5: total number of nodes in tree is 9
Test 6: the largest BST in binary tree is 5
Test 7: traverse node in tree boundary left to right
The tree topology:
10 (L=6, R=4)
15 (L=2, R=3)
16 (L=2, R=1)
17 (L=0, R=0)
-
-
18 (L=0, R=1)
19 (L=0, R=0)
-
-
-
14 (L=1, R=0)
-
13 (L=0, R=0)
-
-
5 (L=5, R=2)
8 (L=0, R=1)
9 (L=0, R=0)
-
-
-
2 (L=0, R=4)
4 (L=3, R=0)
-
3 (L=0, R=2)
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
-
node values: 10 5 2 4 3 6 7 9 13 19 18 17 16 15
Test 8: traverse node in vertical axis left to right
1 (L=1, R=3)
3 (L=2, R=1)
6 (L=0, R=0)
-
-
5 (L=1, R=1)
8 (L=0, R=0)
-
-
7 (L=0, R=0)
-
-
2 (L=0, R=0)
-
-
Nodes values (an axis per row):
2 7
1 5
3 8
6
Test 9: Find the diameter of a binary tree
The tree topology:
1 (L=2, R=3)
3 (L=2, R=1)
6 (L=0, R=0)
-
-
5 (L=1, R=1)
8 (L=0, R=0)
-
-
7 (L=0, R=0)
-
-
2 (L=0, R=1)
4 (L=0, R=0)
-
-
-
The tree diameter is 6
Adjusting the tree, the new topology:
1 (L=0, R=3)
3 (L=2, R=2)
6 (L=0, R=1)
9 (L=0, R=0)
-
-
-
5 (L=1, R=1)
8 (L=0, R=0)
-
-
7 (L=0, R=0)
-
-
-
The tree diameter is 5
Test 10: Print diagonal traversal of a binary tree
The tree topology:
1 (L=2, R=3)
3 (L=2, R=1)
6 (L=0, R=0)
-
-
5 (L=1, R=1)
8 (L=0, R=0)
-
-
7 (L=0, R=0)
-
-
2 (L=1, R=0)
-
4 (L=0, R=0)
-
-
1 3 6
2 5 8
4 7
Test 11: find the distance between two nodes
The distance between 8 and 7 is 2
The distance between 6 and 7 is 3
Test 12: print sum values in diagonal level of tree
10
15
11
Test 13: print the all ancestors of node in reverse order
8 = 5 3 1
1 =
Test 14: determine if a tree is a sum tree
The tree topology:
26 (L=2, R=2)
3 (L=0, R=1)
3 (L=0, R=0)
-
-
-
10 (L=1, R=1)
6 (L=0, R=0)
-
-
4 (L=0, R=0)
-
-
This is a sum tree
Adjust root->left->left->val to 5, the Tree topology:
26 (L=2, R=2)
3 (L=0, R=1)
3 (L=0, R=0)
-
-
-
10 (L=1, R=1)
6 (L=0, R=0)
-
-
5 (L=0, R=0)
-
-
This is not a sum tree
Converting tree into a sum tree, the tree topology:
28 (L=2, R=2)
3 (L=0, R=1)
3 (L=0, R=0)
-
-
-
11 (L=1, R=1)
6 (L=0, R=0)
-
-
5 (L=0, R=0)
-
-
This is a sum tree
run btreebltraverse.out
breadth level traverse (recursive) = 0, 1, 2, 3, 4, 5, 6, 8, 7
breadth level traverse (iterative) = 0, 1, 2, 3, 4, 5, 6, 8, 7
run btreebuild.out
Inorder list: 9, 3, 15, 20, 7
Postorder list: 9, 15, 7, 20, 3
Building binary tree from inorder and postorder list, the tree topology:
3 (L=1, R=2)
20 (L=1, R=1)
7 (L=0, R=0)
-
-
15 (L=0, R=0)
-
-
9 (L=0, R=0)
-
-
Inorder list: 2, 1
Postorder list: 2, 1
Building binary tree from inorder and postorder list, the tree topology:
1 (L=1, R=0)
-
2 (L=0, R=0)
-
-
Inorder list: 1, 2
Postorder list: 2, 1
Building binary tree from inorder and postorder list, the tree topology:
1 (L=0, R=1)
2 (L=0, R=0)
-
-
-
run btreeidentical.out
1st tree topology:
5 (L=2, R=1)
7 (L=0, R=0)
-
-
3 (L=0, R=1)
4 (L=0, R=0)
-
-
-
2nd tree topology:
5 (L=2, R=1)
7 (L=0, R=0)
-
-
3 (L=0, R=1)
4 (L=0, R=0)
-
-
-
Both trees are identical
2nd tree is changed, the tree topology:
5 (L=2, R=2)
7 (L=0, R=1)
9 (L=0, R=0)
-
-
-
3 (L=0, R=1)
4 (L=0, R=0)
-
-
-
Both trees are not identical
run btreeisbalanced.out
Binary tree of adding 3, 2, 4, 5, 6, the tree topology:
3 (L=1, R=3)
4 (L=0, R=2)
5 (L=0, R=1)
6 (L=0, R=0)
-
-
-
-
2 (L=0, R=0)
-
-
The tree is not AVL balanced, rebalancing the tree, the tree topology:
3 (L=1, R=2)
5 (L=1, R=1)
6 (L=0, R=0)
-
-
4 (L=0, R=0)
-
-
2 (L=0, R=0)
-
-
The tree is now AVL balanced
run btreelca.out
The tree topology:
0 (L=2, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
The least common ancestor of 1 and 2 is 0
The least common ancestor of 3 and 4 is 1
The least common ancestor of 2 and 5 is 2
The least common ancestor of 2 and 8 is 2
The least common ancestor of 1 and 8 is 0
The least common ancestor of 3 and 8 is 0
run btreemaxminlevel.out
The tree topology:
0 (L=2, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
Max tree level = 5
Min tree level = 3
run btreemaxnodeinlevel.out
The tree topology:
0 (L=2, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
The maximum number of nodes in a level is 3
Extending branch 3 with two child nodes, the new tree topology:
0 (L=3, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=2, R=1)
4 (L=0, R=0)
-
-
3 (L=1, R=1)
300 (L=0, R=0)
-
-
30 (L=0, R=0)
-
-
The maximum number of nodes in a level is 4
run btreemaxpathsum.out
The tree topology:
0 (L=2, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
The maximum path sum is 20
run btreemaxsumpathbetween2leaves.out
The tree topology:
0 (L=2, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
The maximum sum path between leaves 3 and 4 is 8
The maximum sum path between leaves 8 and 7 is 26
run btreemaxsumpathtoleaf.out
The tree topology:
0 (L=2, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
The maximum sum of path to leaf value 3 is 4
The maximum sum of path to leaf value 4 is 5
The maximum sum of path to leaf value 7 is 20
The maximum sum of path to leaf value 8 is 15
run btreemirrorswap.out
The tree topology:
5 (L=2, R=1)
7 (L=0, R=0)
-
-
3 (L=0, R=1)
4 (L=0, R=0)
-
-
-
The tree topology after mirror swap of tree node:
5 (L=1, R=2)
3 (L=1, R=0)
-
4 (L=0, R=0)
-
-
7 (L=0, R=0)
-
-
run btreepathsum.out
The tree topology:
0 (L=2, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
target sum (15) is 1
target sum (6) is 0
target sum (5) is 1
target sum (7) is 0
run btreerebalancing.out
Binary tree topology of adding 6, 7, 8, 5, 4, 10, 9:
6 (L=2, R=4)
7 (L=0, R=3)
8 (L=0, R=2)
10 (L=1, R=0)
-
9 (L=0, R=0)
-
-
-
-
5 (L=1, R=0)
-
4 (L=0, R=0)
-
-
The binary tree is not AVL balanced
After deleting 8, the tree topology:
6 (L=2, R=3)
7 (L=0, R=2)
9 (L=0, R=1)
10 (L=0, R=0)
-
-
-
-
5 (L=1, R=0)
-
4 (L=0, R=0)
-
-
The binary tree is not AVL balanced
Self balancing binary tree of adding 6, 7, 8, 5, 4, 10, 9, 11:
7 (L=2, R=3)
9 (L=1, R=2)
10 (L=0, R=1)
11 (L=0, R=0)
-
-
-
8 (L=0, R=0)
-
-
5 (L=1, R=1)
6 (L=0, R=0)
-
-
4 (L=0, R=0)
-
-
The binary tree is AVL balanced
After deleting 5 and 6, the tree topology:
9 (L=2, R=2)
10 (L=0, R=1)
11 (L=0, R=0)
-
-
-
7 (L=1, R=1)
8 (L=0, R=0)
-
-
4 (L=0, R=0)
-
-
The binary tree is AVL balanced
run btreesubtree.out
The tree topology:
0 (L=2, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
In order list: 3, 1, 4, 0, 2, 6, 7, 5, 8
Pre order list: 0, 1, 3, 4, 2, 5, 6, 7, 8
Post order list: 3, 4, 1, 7, 6, 8, 5, 2, 0
The 2nd tree topology:
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
In order list: 3, 1, 4
Pre order list: 1, 3, 4
Post order list: 3, 4, 1
The 2nd tree is sub tree of 1st tree
Adjusting 2nd tree by by swapping left and right branch value
The adjusted 2nd tree topology:
1 (L=1, R=1)
3 (L=0, R=0)
-
-
4 (L=0, R=0)
-
-
In order list: 4, 1, 3
Pre order list: 1, 4, 3
Post order list: 4, 3, 1
The 2nd tree is not sub tree of 1st tree
Adjusting 2nd tree by extending val 3 with left child 5
The adjusted 2nd tree topology:
1 (L=2, R=1)
4 (L=0, R=0)
-
-
3 (L=1, R=0)
-
5 (L=0, R=0)
-
-
In order list: 5, 3, 1, 4
Pre order list: 1, 3, 5, 4
Post order list: 5, 3, 4, 1
The 2nd tree is not sub tree of 1st tree
run btreesymmetriccheck.out
The tree topology:
0 (L=2, R=2)
1 (L=1, R=1)
2 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
1 (L=1, R=1)
3 (L=0, R=0)
-
-
2 (L=0, R=0)
-
-
The tree is symmetric
The tree topology (after setting root->right->left to NULL):
0 (L=2, R=2)
1 (L=0, R=1)
2 (L=0, R=0)
-
-
-
1 (L=1, R=1)
3 (L=0, R=0)
-
-
2 (L=0, R=0)
-
-
The tree is not symmetric
run btreetraverse.out
The tree topology:
0 (L=2, R=4)
2 (L=0, R=3)
5 (L=2, R=1)
8 (L=0, R=0)
-
-
6 (L=0, R=1)
7 (L=0, R=0)
-
-
-
-
1 (L=1, R=1)
4 (L=0, R=0)
-
-
3 (L=0, R=0)
-
-
pre order = 0, 1, 3, 4, 2, 5, 6, 7, 8
post order = 3, 4, 1, 7, 6, 8, 5, 2, 0
in order = 3, 1, 4, 0, 2, 6, 7, 5, 8
run btreeverticalsum.out
The tree topology:
1 (L=1, R=3)
3 (L=2, R=1)
6 (L=0, R=0)
-
-
5 (L=1, R=1)
8 (L=0, R=0)
-
-
7 (L=0, R=0)
-
-
2 (L=0, R=0)
-
-
The tree in aligned vertical axis
1
2 3
5 6
7 8
Vertical node sum:
9 6 11 6
run tree2btreebuild.out
Nary tree topology:
1
30
20
10
102
101
100
Build a binary search tree, the tree topology:
100 (L=3, R=2)
101 (L=0, R=1)
102 (L=0, R=0)
-
-
-
10 (L=1, R=2)
20 (L=0, R=1)
30 (L=0, R=0)
-
-
-
1 (L=0, R=0)
-
-