diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md index 36b39740..1b777dd5 100644 --- a/problems/0101.对称二叉树.md +++ b/problems/0101.对称二叉树.md @@ -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(); + 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; +} +```

diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 51d030e7..e977279c 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -462,6 +462,32 @@ impl Solution { } } ``` +### C# +```C# +public IList> LevelOrder(TreeNode root) +{ + var res = new List>(); + var que = new Queue(); + if (root == null) return res; + que.Enqueue(root); + while (que.Count != 0) + { + var size = que.Count; + var vec = new List(); + 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> LevelOrderBottom(TreeNode root) +{ + var res = new List>(); + var que = new Queue(); + if (root == null) return res; + que.Enqueue(root); + while (que.Count != 0) + { + var size = que.Count; + var vec = new List(); + 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 diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md index 05044375..ab3ce52e 100644 --- a/problems/0104.二叉树的最大深度.md +++ b/problems/0104.二叉树的最大深度.md @@ -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 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; +} +```

diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md index 11517783..8691953a 100644 --- a/problems/0226.翻转二叉树.md +++ b/problems/0226.翻转二叉树.md @@ -1018,25 +1018,13 @@ public class Solution { ```csharp //迭代 public class Solution { - public TreeNode InvertTree(TreeNode root) { - if (root == null) return null; - Stack stack=new Stack(); - 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); - } - return root; - } - - public void swap(TreeNode node) { - TreeNode temp = node.left; - node.left = node.right; - node.right = temp; - } +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; +} } ``` diff --git a/problems/二叉树的统一迭代法.md b/problems/二叉树的统一迭代法.md index 8089af64..ad79424d 100644 --- a/problems/二叉树的统一迭代法.md +++ b/problems/二叉树的统一迭代法.md @@ -741,6 +741,99 @@ impl Solution{ } } ``` +### C# +```C# +// 前序遍历 +public IList PreorderTraversal(TreeNode root) +{ + var res = new List(); + var st = new Stack(); + 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 InorderTraversal(TreeNode root) +{ + var res = new List(); + var st = new Stack(); + 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 PostorderTraversal(TreeNode root) +{ + var res = new List(); + var st = new Stack(); + 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; +} +```

diff --git a/problems/二叉树的迭代遍历.md b/problems/二叉树的迭代遍历.md index 69007995..e39709b8 100644 --- a/problems/二叉树的迭代遍历.md +++ b/problems/二叉树的迭代遍历.md @@ -695,6 +695,67 @@ impl Solution { } } ``` +### C# +```C# +// 前序遍历 +public IList PreorderTraversal(TreeNode root) +{ + var st = new Stack(); + var res = new List(); + 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 InorderTraversal(TreeNode root) +{ + var st = new Stack(); + var res = new List(); + 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 PostorderTraversal(TreeNode root) +{ + var res = new List(); + var st = new Stack(); + 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; +} +```

diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md index 730b18ac..6dad6e56 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -565,6 +565,60 @@ impl Solution { } ``` +### C# +```C# +// 前序遍历 +public IList PreorderTraversal(TreeNode root) +{ + var res = new List(); + if (root == null) return res; + Traversal(root, res); + return res; + +} +public void Traversal(TreeNode cur, IList res) +{ + if (cur == null) return; + res.Add(cur.val); + Traversal(cur.left, res); + Traversal(cur.right, res); +} +``` +```C# +// 中序遍历 +public IList InorderTraversal(TreeNode root) +{ + var res = new List(); + if (root == null) return res; + Traversal(root, res); + return res; +} +public void Traversal(TreeNode cur, IList res) +{ + if (cur == null) return; + Traversal(cur.left, res); + res.Add(cur.val); + Traversal(cur.right, res); +} +``` +```C# +// 后序遍历 +public IList PostorderTraversal(TreeNode root) +{ + var res = new List(); + if (root == null) return res; + Traversal(root, res); + return res; +} +public void Traversal(TreeNode cur, IList res) +{ + if (cur == null) return; + Traversal(cur.left, res); + Traversal(cur.right, res); + res.Add(cur.val); +} +``` +