添加（0226.翻转二叉树.md）：增加typescript版本

problems/0226.翻转二叉树.md

@@ -563,7 +563,135 @@ var invertTree = function(root) {
 };
 ```
 
+### TypeScript：
+
+递归法：
+
+```typescript
+// 递归法（前序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    if (root === null) return root;
+    let tempNode: TreeNode | null = root.left;
+    root.left = root.right;
+    root.right = tempNode;
+    invertTree(root.left);
+    invertTree(root.right);
+    return root;
+};
+
+// 递归法（后序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    if (root === null) return root;
+    invertTree(root.left);
+    invertTree(root.right);
+    let tempNode: TreeNode | null = root.left;
+    root.left = root.right;
+    root.right = tempNode;
+    return root;
+};
+
+// 递归法（中序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    if (root === null) return root;
+    invertTree(root.left);
+    let tempNode: TreeNode | null = root.left;
+    root.left = root.right;
+    root.right = tempNode;
+    // 因为左右节点已经进行交换，此时的root.left 是原先的root.right
+    invertTree(root.left);
+    return root;
+};
+```
+
+迭代法：
+
+```typescript
+// 迭代法（栈模拟前序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    let helperStack: TreeNode[] = [];
+    let curNode: TreeNode,
+        tempNode: TreeNode | null;
+    if (root !== null) helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop()!;
+        // 入栈操作最好在交换节点之前进行，便于理解
+        if (curNode.right) helperStack.push(curNode.right);
+        if (curNode.left) helperStack.push(curNode.left);
+        tempNode = curNode.left;
+        curNode.left = curNode.right;
+        curNode.right = tempNode;
+    }
+    return root;
+};
+
+// 迭代法（栈模拟中序遍历-统一写法形式）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    let helperStack: (TreeNode | null)[] = [];
+    let curNode: TreeNode | null,
+        tempNode: TreeNode | null;
+    if (root !== null) helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop();
+        if (curNode !== null) {
+            if (curNode.right !== null) helperStack.push(curNode.right);
+            helperStack.push(curNode);
+            helperStack.push(null);
+            if (curNode.left !== null) helperStack.push(curNode.left);
+        } else {
+            curNode = helperStack.pop()!;
+            tempNode = curNode.left;
+            curNode.left = curNode.right;
+            curNode.right = tempNode;
+        }
+    }
+    return root;
+};
+
+// 迭代法（栈模拟后序遍历-统一写法形式）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    let helperStack: (TreeNode | null)[] = [];
+    let curNode: TreeNode | null,
+        tempNode: TreeNode | null;
+    if (root !== null) helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop();
+        if (curNode !== null) {
+            helperStack.push(curNode);
+            helperStack.push(null);
+            if (curNode.right !== null) helperStack.push(curNode.right);
+            if (curNode.left !== null) helperStack.push(curNode.left);
+        } else {
+            curNode = helperStack.pop()!;
+            tempNode = curNode.left;
+            curNode.left = curNode.right;
+            curNode.right = tempNode;
+        }
+    }
+    return root;
+};
+
+// 迭代法（队列模拟层序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    const helperQueue: TreeNode[] = [];
+    let curNode: TreeNode,
+        tempNode: TreeNode | null;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            curNode = helperQueue.shift()!;
+            tempNode = curNode.left;
+            curNode.left = curNode.right;
+            curNode.right = tempNode;
+            if (curNode.left !== null) helperQueue.push(curNode.left);
+            if (curNode.right !== null) helperQueue.push(curNode.right);
+        }
+    }
+    return root;
+};
+```
+
 ### C:
+
 递归法
 ```c
 struct TreeNode* invertTree(struct TreeNode* root){