/*
Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
*/
public Boolean IsSameTree(TreeNode p, TreeNode q)
{
if (p == null && q == null)
{
return true;
}
if (p == null || q == null)
{
return false;
}
if (p.val == q.val)
{
return IsSameTree(p.left, q.left) && IsSameTree(p.right, q.right);
}
return false;
}
private static void IsSymmetric()
{
TreeNode a = new TreeNode(1);
a.left = new TreeNode(2);
a.left.right = new TreeNode(3);
a.right = new TreeNode(2);
a.right.right = new TreeNode(3);
TreeNode b = new TreeNode(1);
b.left = new TreeNode(2);
b.right = new TreeNode(2);
var aa = new LeetCodeOJ154.Solution();
Console.WriteLine(aa.IsSymmetric(a));
Console.WriteLine(aa.IsSymmetric(b));
}
private static void IsSameTree()
{
TreeNode a = new TreeNode(1);
a.left = new TreeNode(2);
a.right = new TreeNode(3);
a.right.left = new TreeNode(5);
TreeNode b = new TreeNode(1);
b.left = new TreeNode(2);
b.right = new TreeNode(4);
var aa = new LeetCodeOJ154.Solution();
Console.WriteLine(aa.IsSameTree(a,b));
Console.ReadKey();
}
private Boolean IsSymmetric2(TreeNode p, TreeNode q)
{
if (p == null && q == null)
{
return true;
}
if (p == null || q == null)
{
return false;
}
if (p.val == q.val)
{
return IsSymmetric2(p.left, q.right) && IsSymmetric2(p.right, q.left);
}
return false;
}
public IList<IList<int>> LevelOrder2(TreeNode root)
{
IList<IList<int>> result = new List<IList<int>>();
if (root == null)
{
return result;
}
return result;
}
/*
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
*/
public IList<IList<int>> LevelOrder(TreeNode root)
{
var result = new List<IList<int>>();
if (root == null)
{
return result;
}
var queue = new Queue<KeyValuePair<TreeNode, int>>();
queue.Enqueue(new KeyValuePair<TreeNode, int>(root, 0));
while (queue.Count > 0)
{
var kvp = queue.Dequeue();
var node = kvp.Key;
var level = kvp.Value;
if (level >= result.Count)
{
result.Add(new List<int>());
}
result[level].Add(node.val);
if (node.left != null) queue.Enqueue(new KeyValuePair<TreeNode, int>(node.left, level + 1));
if (node.right != null) queue.Enqueue(new KeyValuePair<TreeNode, int>(node.right, level + 1));
}
return result;
}
/*
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
*/
public Boolean IsSymmetric(TreeNode root)
{
if (root == null)
{
return true;
}
return IsSymmetric2(root.left, root.right);
}
