ConstructBinaryTreefromPreorderandInorderTraversal

Change-Id: Ifde5ecad675fc6b7f00fcf9d41ed414edb3547bb

Author: applewjg

ConstructBinaryTreefromPreorderandInorderTraversal.java

@@ -0,0 +1,43 @@
+/*
+ Author:     Andy, nkuwjg@gmail.com
+ Date:       Jan 16, 2015
+ Problem:    Construct Binary Tree from Preorder and Inorder Traversal
+ Difficulty: Medium
+ Source:     https://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
+ Notes:
+ Given preorder and inorder traversal of a tree, construct the binary tree.
+ Note:
+ You may assume that duplicates do not exist in the tree.
+
+ Solution: Recursion.
+ */
+
+/**
+ * Definition for binary tree
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+public class Solution {
+    public TreeNode buildTree(int[] preorder, int[] inorder) {
+        if(inorder.length == 0 || preorder.length == 0 || inorder.length != preorder.length) return null;
+        return buildTreeRe(preorder, 0, inorder, 0, preorder.length);
+    }
+    public TreeNode buildTreeRe(int[] preorder, int s1, int[] inorder, int s2, int size) {
+        if (size <= 0 ) return null;
+        TreeNode node = new TreeNode(preorder[s1]);
+        if (size == 1) return node;
+        int pos = s2;
+        while (pos <= (s2 + size - 1)) {
+            if (inorder[pos] == preorder[s1]) break;
+            ++pos;
+        }
+        int leftlen = pos - s2;
+        node.left = buildTreeRe(preorder, s1 + 1, inorder, s2, leftlen);
+        node.right = buildTreeRe(preorder, s1 + leftlen + 1, inorder, pos + 1, size - leftlen - 1);
+        return node;
+    }
+}