feat: solve construct binary tree from prorder and inorder traversal

Author: GangBean

construct-binary-tree-from-preorder-and-inorder-traversal/GangBean.java

@@ -0,0 +1,51 @@
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode() {}
+ *     TreeNode(int val) { this.val = val; }
+ *     TreeNode(int val, TreeNode left, TreeNode right) {
+ *         this.val = val;
+ *         this.left = left;
+ *         this.right = right;
+ *     }
+ * }
+ */
+class Solution {
+    public TreeNode buildTree(int[] preorder, int[] inorder) {
+        /**
+            1. understanding
+            - preorder: mid -> left -> right
+            - inorder: left -> mid -> right
+            - so, first element of the preorder array is always mid node.
+            - if the idx of inorder's 1st depth mid node is k, then inorder[0:k-1] is the left tree part array. And also, preorder[k:] is the right tree part.
+            2. strategy
+            - find the inorder's mid node idx, and then split left tree part and right part, buildTree with each preorder and inorder part.
+
+            3. complexity
+            - time: O(N^2) 
+            - space: O(N^2)
+        */
+        if (preorder.length == 0) return null;
+        if (preorder.length == 1) return new TreeNode(preorder[0]);
+        int i = 0;
+        List<Integer> leftPreorder = new ArrayList<>(); // O(N)
+        List<Integer> leftInorder = new ArrayList<>(); // O(N)
+        List<Integer> rightPreorder = new ArrayList<>(); // O(N)
+        List<Integer> rightInorder = new ArrayList<>(); // O(N)
+        for (; i < inorder.length; i++) { // O(N)
+            if (inorder[i] == preorder[0]) break;
+            leftPreorder.add(preorder[i+1]);
+            leftInorder.add(inorder[i]);
+        }
+        for (int idx = i+1; idx < inorder.length; idx++) { // O(N)
+            rightPreorder.add(preorder[idx]);
+            rightInorder.add(inorder[idx]);
+        }
+
+        return new TreeNode(preorder[0], buildTree(leftPreorder.stream().mapToInt(Integer::intValue).toArray(), leftInorder.stream().mapToInt(Integer::intValue).toArray()), buildTree(rightPreorder.stream().mapToInt(Integer::intValue).toArray(), rightInorder.stream().mapToInt(Integer::intValue).toArray()));
+    }
+}
+