CODE 27: Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

	public TreeNode buildTree(int[] inorder, int[] postorder) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if (null == inorder || null == postorder || inorder.length <= 0
				|| postorder.length <= 0) {
			return null;
		}
		return dfs(inorder, postorder, 0, inorder.length - 1, 0,
				postorder.length - 1);
	}

	TreeNode dfs(int[] inorder, int[] postorder, int inorderStart,
			int inorderEnd, int postorderStart, int postorderEnd) {
		TreeNode node = new TreeNode(postorder[postorderEnd]);
		int nodeIndexInInorder = findElement(inorder, postorder[postorderEnd]);
		int leftLength = nodeIndexInInorder - inorderStart;
		int rightLength = inorderEnd - nodeIndexInInorder;
		if (leftLength > 0) {
			TreeNode left = dfs(inorder, postorder, inorderStart,
					nodeIndexInInorder - 1, postorderStart, postorderEnd
							- rightLength - 1);
			node.left = left;
		}
		if (rightLength > 0) {
			TreeNode right = dfs(inorder, postorder, nodeIndexInInorder + 1,
					inorderEnd, postorderEnd - rightLength, postorderEnd - 1);
			node.right = right;
		}
		return node;
	}

	int findElement(int[] array, int element) {
		for (int i = 0; i < array.length; i++) {
			if (array[i] == element) {
				return i;
			}
		}
		return -1;
	}