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Author: BaffinLee

101-200/105. Construct Binary Tree from Preorder and Inorder Traversal.md

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+# 105. Construct Binary Tree from Preorder and Inorder Traversal
+
+- Difficulty: Medium.
+- Related Topics: Array, Tree, Depth-first Search.
+- Similar Questions: Construct Binary Tree from Inorder and Postorder Traversal.
+
+## Problem
+
+Given preorder and inorder traversal of a tree, construct the binary tree.
+
+**Note:**
+You may assume that duplicates do not exist in the tree.
+
+For example, given
+
+```
+preorder = [3,9,20,15,7]
+inorder = [9,3,15,20,7]
+```
+
+Return the following binary tree:
+
+```
+    3
+   / \
+  9  20
+    /  \
+   15   7
+```
+
+## Solution
+
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {number[]} preorder
+ * @param {number[]} inorder
+ * @return {TreeNode}
+ */
+var buildTree = function(preorder, inorder) {
+  return helper(preorder, inorder, 0, 0, inorder.length - 1);
+};
+
+var helper = function (preorder, inorder, preIndex, inStart, inEnd) {
+  if (preIndex >= preorder.length || inStart > inEnd) return null;
+  
+  var index = 0;
+  var root = new TreeNode(preorder[preIndex]);
+  
+  for (var i = inStart; i <= inEnd; i++) {
+    if (inorder[i] === root.val) {
+      index = i;
+      break;
+    }
+  }
+  
+  if (index > inStart) root.left = helper(preorder, inorder, preIndex + 1, inStart, index - 1);
+  if (index < inEnd) root.right = helper(preorder, inorder, preIndex + index - inStart + 1, index + 1, inEnd);
+  
+  return root;
+};
+```
+
+**Explain:**
+
+1. `preorder[0]` 是 `root`
+2. 在 `inorder` 里找到 `root`，左边是 `root.left`，右边是 `root.right`。
+3. 递归
+
+**Complexity:**
+
+* Time complexity :
+* Space complexity :

101-200/106. Construct Binary Tree from Inorder and Postorder Traversal.md

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+# 106. Construct Binary Tree from Inorder and Postorder Traversal
+
+- Difficulty: Medium.
+- Related Topics: Array, Tree, Depth-first Search.
+- Similar Questions: Construct Binary Tree from Preorder and Inorder Traversal.
+
+## Problem
+
+Given inorder and postorder traversal of a tree, construct the binary tree.
+
+**Note:**
+
+You may assume that duplicates do not exist in the tree.
+
+For example, given
+
+```
+inorder = [9,3,15,20,7]
+postorder = [9,15,7,20,3]
+```
+
+Return the following binary tree:
+
+```
+    3
+   / \
+  9  20
+    /  \
+   15   7
+```
+
+## Solution
+
+```javascript
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {number[]} inorder
+ * @param {number[]} postorder
+ * @return {TreeNode}
+ */
+var buildTree = function(inorder, postorder) {
+  return helper(inorder, postorder, 0, inorder.length - 1, postorder.length - 1);
+};
+
+var helper = function (inorder, postorder, inStart, inEnd, postIndex) {
+  if (inStart > inEnd || postIndex < 0) return null;
+  
+  var index = 0;
+  var root = new TreeNode(postorder[postIndex]);
+  
+  for (var i = inStart; i <= inEnd; i++) {
+    if (inorder[i] === root.val) {
+      index = i;
+      break;
+    }
+  }
+  
+  root.right = helper(inorder, postorder, index + 1, inEnd, postIndex - 1);
+  root.left = helper(inorder, postorder, inStart, index - 1, postIndex - 1 - (inEnd - index));
+  
+  return root;
+};
+```
+
+**Explain:**
+
+see [Construct Binary Tree from Preorder and Inorder Traversal](./construct-binary-tree-from-preorder-and-inorder-traversal.html).
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).