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Merge pull request #2334 from eeee0717/master

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Update.二叉树的递归遍历，添加C#版
This commit is contained in:
程序员Carl
2023-11-16 12:08:14 +08:00
committed by GitHub
parent 2a8c571391 3ffca338ca
commit 3334e5fc09
7 changed files with 368 additions and 19 deletions
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47
problems/0101.对称二叉树.md
View File

@ -897,6 +897,53 @@ impl Solution {
}
}
```
### C#
```C#
// 递归
public bool IsSymmetric(TreeNode root)
{
if (root == null) return true;
return Compare(root.left, root.right);
}
public bool Compare(TreeNode left, TreeNode right)
{
if(left == null && right != null) return false;
else if(left != null && right == null ) return false;
else if(left == null && right == null) return true;
else if(left.val != right.val) return false;
var outside = Compare(left.left, right.right);
var inside = Compare(left.right, right.left);
return outside&&inside;
}
```
``` C#
// 迭代法
public bool IsSymmetric(TreeNode root)
{
if (root == null) return true;
var st = new Stack<TreeNode>();
st.Push(root.left);
st.Push(root.right);
while (st.Count != 0)
{
var left = st.Pop();
var right = st.Pop();
if (left == null && right == null)
continue;
if ((left == null || right == null || (left.val != right.val)))
return false;
st.Push(left.left);
st.Push(right.right);
st.Push(left.right);
st.Push(right.left);
}
return true;
}
```
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51
problems/0102.二叉树的层序遍历.md
View File

@ -462,6 +462,32 @@ impl Solution {
}
}
```
### C#
```C#
public IList<IList<int>> LevelOrder(TreeNode root)
{
var res = new List<IList<int>>();
var que = new Queue<TreeNode>();
if (root == null) return res;
que.Enqueue(root);
while (que.Count != 0)
{
var size = que.Count;
var vec = new List<int>();
for (int i = 0; i < size; i++)
{
var cur = que.Dequeue();
vec.Add(cur.val);
if (cur.left != null) que.Enqueue(cur.left);
if (cur.right != null) que.Enqueue(cur.right);
}
res.Add(vec);
}
return res;
}
```
**此时我们就掌握了二叉树的层序遍历了，那么如下九道力扣上的题目，只需要修改模板的两三行代码（不能再多了），便可打倒！**
@ -798,6 +824,31 @@ impl Solution {
}
}
```
### C#
```C#
public IList<IList<int>> LevelOrderBottom(TreeNode root)
{
var res = new List<IList<int>>();
var que = new Queue<TreeNode>();
if (root == null) return res;
que.Enqueue(root);
while (que.Count != 0)
{
var size = que.Count;
var vec = new List<int>();
for (int i = 0; i < size; i++)
{
var cur = que.Dequeue();
vec.Add(cur.val);
if (cur.left != null) que.Enqueue(cur.left);
if (cur.right != null) que.Enqueue(cur.right);
}
res.Add(vec);
}
res.Reverse();
return res;
}
```
## 199.二叉树的右视图

55
problems/0104.二叉树的最大深度.md
View File

@ -1022,6 +1022,61 @@ impl Solution {
max_depth
}
```
### C#
```C#
// 递归法
public int MaxDepth(TreeNode root) {
if(root == null) return 0;
int leftDepth = MaxDepth(root.left);
int rightDepth = MaxDepth(root.right);
return 1 + Math.Max(leftDepth, rightDepth);
}
```
```C#
// 前序遍历
int result = 0;
public int MaxDepth(TreeNode root)
{
if (root == null) return result;
GetDepth(root, 1);
return result;
}
public void GetDepth(TreeNode root, int depth)
{
result = depth > result ? depth : result;
if (root.left == null && root.right == null) return;
if (root.left != null)
GetDepth(root.left, depth + 1);
if (root.right != null)
GetDepth(root.right, depth + 1);
return;
}
```
```C#
// 迭代法
public int MaxDepth(TreeNode root)
{
int depth = 0;
Queue<TreeNode> que = new();
if (root == null) return depth;
que.Enqueue(root);
while (que.Count != 0)
{
int size = que.Count;
depth++;
for (int i = 0; i < size; i++)
{
var node = que.Dequeue();
if (node.left != null) que.Enqueue(node.left);
if (node.right != null) que.Enqueue(node.right);
}
}
return depth;
}
```
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24
problems/0226.翻转二叉树.md
View File

@ -1018,25 +1018,13 @@ public class Solution {
```csharp
//迭代
public class Solution {
public TreeNode InvertTree(TreeNode root) {
if (root == null) return null;
Stack<TreeNode> stack=new Stack<TreeNode>();
stack.Push(root);
while(stack.Count>0)
{
TreeNode node = stack.Pop();
swap(node);
if(node.right!=null) stack.Push(node.right);
if(node.left!=null) stack.Push(node.left);
}
public TreeNode InvertTree(TreeNode root) {
if(root == null) return root;
(root.left,root.right) = (root.right, root.left);
InvertTree(root.left);
InvertTree(root.right);
return root;
}
public void swap(TreeNode node) {
TreeNode temp = node.left;
node.left = node.right;
node.right = temp;
}
}
}
```

93
problems/二叉树的统一迭代法.md
View File

@ -741,6 +741,99 @@ impl Solution{
}
}
```
### C#
```C#
// 前序遍历
public IList<int> PreorderTraversal(TreeNode root)
{
var res = new List<int>();
var st = new Stack<TreeNode>();
if (root == null) return res;
st.Push(root);
while (st.Count != 0)
{
var node = st.Peek();
if (node == null)
{
st.Pop();
node = st.Peek();
st.Pop();
res.Add(node.val);
}
else
{
st.Pop();
if (node.right != null) st.Push(node.right);
if (node.left != null) st.Push(node.left);
st.Push(node);
st.Push(null);
}
}
return res;
}
```
```C#
// 中序遍历
public IList<int> InorderTraversal(TreeNode root)
{
var res = new List<int>();
var st = new Stack<TreeNode>();
if (root == null) return res;
st.Push(root);
while (st.Count != 0)
{
var node = st.Peek();
if (node == null)
{
st.Pop();
node = st.Peek();
st.Pop();
res.Add(node.val);
}
else
{
st.Pop();
if (node.right != null) st.Push(node.right);
st.Push(node);
st.Push(null);
if (node.left != null) st.Push(node.left);
}
}
return res;
}
```
```C#
// 后序遍历
public IList<int> PostorderTraversal(TreeNode root)
{
var res = new List<int>();
var st = new Stack<TreeNode>();
if (root == null) return res;
st.Push(root);
while (st.Count != 0)
{
var node = st.Peek();
if (node == null)
{
st.Pop();
node = st.Peek();
st.Pop();
res.Add(node.val);
}
else
{
st.Pop();
if (node.left != null) st.Push(node.left);
if (node.right != null) st.Push(node.right);
st.Push(node);
st.Push(null);
}
}
res.Reverse(0, res.Count);
return res;
}
```
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61
problems/二叉树的迭代遍历.md
View File

@ -695,6 +695,67 @@ impl Solution {
}
}
```
### C#
```C#
// 前序遍历
public IList<int> PreorderTraversal(TreeNode root)
{
var st = new Stack<TreeNode>();
var res = new List<int>();
if (root == null) return res;
st.Push(root);
while (st.Count != 0)
{
var node = st.Pop();
res.Add(node.val);
if (node.right != null)
st.Push(node.right);
if (node.left != null)
st.Push(node.left);
}
return res;
}
// 中序遍历
public IList<int> InorderTraversal(TreeNode root)
{
var st = new Stack<TreeNode>();
var res = new List<int>();
var cur = root;
while (st.Count != 0 || cur != null)
{
if (cur != null)
{
st.Push(cur);
cur = cur.left;
}
else
{
cur = st.Pop();
res.Add(cur.val);
cur = cur.right;
}
}
return res;
}
// 后序遍历
public IList<int> PostorderTraversal(TreeNode root)
{
var res = new List<int>();
var st = new Stack<TreeNode>();
if (root == null) return res;
st.Push(root);
while (st.Count != 0)
{
var cur = st.Pop();
res.Add(cur.val);
if (cur.left != null) st.Push(cur.left);
if (cur.right != null) st.Push(cur.right);
}
res.Reverse(0, res.Count());
return res;
}
```
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54
problems/二叉树的递归遍历.md
View File

@ -565,6 +565,60 @@ impl Solution {
}
```
### C#
```C#
// 前序遍历
public IList<int> PreorderTraversal(TreeNode root)
{
var res = new List<int>();
if (root == null) return res;
Traversal(root, res);
return res;
}
public void Traversal(TreeNode cur, IList<int> res)
{
if (cur == null) return;
res.Add(cur.val);
Traversal(cur.left, res);
Traversal(cur.right, res);
}
```
```C#
// 中序遍历
public IList<int> InorderTraversal(TreeNode root)
{
var res = new List<int>();
if (root == null) return res;
Traversal(root, res);
return res;
}
public void Traversal(TreeNode cur, IList<int> res)
{
if (cur == null) return;
Traversal(cur.left, res);
res.Add(cur.val);
Traversal(cur.right, res);
}
```
```C#
// 后序遍历
public IList<int> PostorderTraversal(TreeNode root)
{
var res = new List<int>();
if (root == null) return res;
Traversal(root, res);
return res;
}
public void Traversal(TreeNode cur, IList<int> res)
{
if (cur == null) return;
Traversal(cur.left, res);
Traversal(cur.right, res);
res.Add(cur.val);
}
```
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