feat: 105. Construct Binary Tree from Preorder and Inorder Traversal

Author: HC-kang

construct-binary-tree-from-preorder-and-inorder-traversal/HC-kang.ts

@@ -0,0 +1,53 @@
+// Definition for a binary tree node.
+class TreeNode {
+  val: number;
+  left: TreeNode | null;
+  right: TreeNode | null;
+  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
+    this.val = val === undefined ? 0 : val;
+    this.left = left === undefined ? null : left;
+    this.right = right === undefined ? null : right;
+  }
+}
+
+// T.C: O(N)
+// S.C: O(N)
+function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
+  if (preorder.length === 0 || inorder.length === 0) {
+    return null;
+  }
+  const root = new TreeNode(preorder[0]);
+  const idx = inorder.indexOf(preorder[0]);
+  root.left = buildTree(preorder.slice(1, idx + 1), inorder.slice(0, idx));
+  root.right = buildTree(preorder.slice(idx + 1), inorder.slice(idx + 1));
+
+  return root;
+}
+
+// Not using slice. but I think it's not necessary... first solution is more readable. and that's not so bad.
+// T.C: O(N)
+// S.C: O(N)
+function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
+  // this tree is consist of unique values
+  const inorderMap = new Map<number, number>();
+  for (const [i, val] of inorder.entries()) {
+    inorderMap.set(val, i);
+  }
+
+  function helper(preLeft: number, preRight: number, inLeft: number, inRight: number): TreeNode | null {
+    if (preLeft > preRight) return null;
+
+    const rootValue = preorder[preLeft];
+    const root = new TreeNode(rootValue);
+    const inRootIdx = inorderMap.get(rootValue)!;
+
+    const leftSize = inRootIdx - inLeft;
+
+    root.left = helper(preLeft + 1, preLeft + leftSize, inLeft, inRootIdx - 1);
+    root.right = helper(preLeft + leftSize + 1, preRight, inRootIdx + 1, inRight);
+
+    return root;
+  }
+
+  return helper(0, preorder.length - 1, 0, inorder.length - 1);
+}