Merge pull request #131 from qilingit/day24

Add Java solution for day24: Construct Binary Search Tree from Preorder Traversal

Author: jinshendan

May-LeetCoding-Challenge/24-Construct-Binary-Search-Tree-From-Preorder-Traversal/Construct-Binary-Search-Tree-From-Preorder-Traversal.java

@@ -0,0 +1,58 @@
+/**
+ * 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 {
+    /**
+     * Recursive solution, construct the left tree and right tree recursively, and the combine to the root.
+     * @param preorder
+     * @return
+     */
+    public TreeNode bstFromPreorder(int[] preorder) {
+        if(preorder.length == 0) return null;
+        
+        int rootVal = preorder[0];
+        if(preorder.length == 1) return new TreeNode(rootVal);
+        
+        TreeNode rightTree = null;
+        TreeNode leftTree = null;
+        TreeNode root = new TreeNode(rootVal, null, null);
+        
+        if(preorder[1] > rootVal) {
+            // if there is no left tree
+            rightTree = bstFromPreorder(Arrays.copyOfRange(preorder, 1, preorder.length));
+        } else {
+            int i = 1;
+            while (i < preorder.length) {
+                // find separate index of left tree and right tree
+                if (preorder[i] > rootVal) {
+                    break;
+                }
+                i++;
+            }
+            
+            // construct left tree
+            leftTree = bstFromPreorder(Arrays.copyOfRange(preorder, 1, i));
+            if(i != preorder.length) {
+                //construct right tree
+                rightTree = bstFromPreorder(Arrays.copyOfRange(preorder, i, preorder.length));
+            }
+        }        
+        
+        root.left = leftTree;
+        root.right = rightTree;
+        
+        return root;
+    }
+}