Solution: Construct Binary Tree from Preorder and Inorder Traversal

Author: obzva

construct-binary-tree-from-preorder-and-inorder-traversal/flynn.cpp

@@ -0,0 +1,49 @@
+/**
+ * For the number of given nodes N,
+ * 
+ * Time complexity: O(N)
+ * 
+ * Space complexity: O(N) at worst
+ */
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
+        unordered_map<int, int> inorder_index_map;
+        stack<TreeNode*> tree_stack;
+
+        for (int i = 0; i < inorder.size(); i++) inorder_index_map[inorder[i]] = i;
+
+        TreeNode* root = new TreeNode(preorder[0]);
+        tree_stack.push(root);
+
+        for (int i = 1; i < preorder.size(); i++) {
+            TreeNode* curr = new TreeNode(preorder[i]);
+
+            if (inorder_index_map[curr->val] < inorder_index_map[tree_stack.top()->val]) {
+                tree_stack.top()->left = curr;
+            } else {
+                TreeNode* parent;
+                while (!tree_stack.empty() && inorder_index_map[curr->val] > inorder_index_map[tree_stack.top()->val]) {
+                    parent = tree_stack.top();
+                    tree_stack.pop();
+                }
+                parent->right = curr;
+            }
+            tree_stack.push(curr);
+        }
+
+        return root;
+    }
+};