Completed Trees-1

ConstructBinaryTreeFromPre-InOrderTraversal.java

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+// Time Complexity : O(n)
+// Space Complexity : O(n)
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+// Your code here along with comments explaining your approach
+/*
+Maintain a map to store the inorder array elements and their indices to understand the left and right elements
+of root, once we identify the root from preorder array. We can have a helper method to get root element using
+a pointer for each recursion and leverage that root element and map to find root's index. All the values before
+that index in inorder array would be left and after the index is right. This way, we can construct the whole tree.
+ */
+/**
+ * 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 {
+    int index;
+    public TreeNode buildTree(int[] preorder, int[] inorder) {
+       this.index = 0;
+       Map<Integer, Integer> map = new HashMap<>();
+       for(int i = 0 ; i < inorder.length ; i++)
+           map.put(inorder[i], i);
+       return helper(preorder, 0, preorder.length - 1, map);
+    }
+
+    private TreeNode helper(int[] preorder, int start, int end, Map<Integer, Integer> map) {
+        if(start > end)
+            return null;
+        int rootVal = preorder[index];
+        index++;
+
+        int rootIndex = map.get(rootVal);
+        TreeNode root = new TreeNode(rootVal);
+
+        root.left = helper(preorder, start, rootIndex - 1, map);
+        root.right = helper(preorder, rootIndex +  1, end, map);
+        return root;
+    }
+}

ValidateBST.java

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+// Time Complexity : O(n)
+// Space Complexity : O(h) - height of the tree
+// Did this code successfully run on Leetcode : yes
+// Any problem you faced while coding this : no
+
+// Your code here along with comments explaining your approach
+/*
+We use a helper method to traverse left and right children of root recursively and make sure to check
+if the previous visited node's value is not greater than root value at each stack call, if so, update the
+flag as false.We need to update the previous visited node with the current root to check for next stack/
+child calls. This concept/intention maps to inorder traversal where visited elements are to be in ascending
+order.
+ */
+/**
+ * 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 {
+    TreeNode prev;
+    boolean flag;
+
+    public boolean isValidBST(TreeNode root) {
+        this.flag = true;
+        validateBST(root);
+        return flag;
+    }
+
+    private void validateBST(TreeNode root) {
+        if(root == null)
+            return;
+        validateBST(root.left);
+        if(prev != null && prev.val >= root.val)
+            flag = false;
+        prev = root;
+        validateBST(root.right);
+    }
+}