LeetCode:Construct Binary Tree from Preorder and Inorder Traversal

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

Note:

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

以前在某ACM网站做过这个题，没用递归，差点死了。

这题用递归实现很简单。

public class ConstructBinaryTreefromPreorderandInorderTraversal {
	
	public TreeNode buildTree(int[] preorder, int[] inorder) {
		if (preorder == null || preorder.length == 0) {
			return null;
		}
        return buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
    }
	
	public TreeNode buildTree(int[] preorder, int start, int end, int[] inorder, int inStart, int inEnd) {
		int rootValue = preorder[start];
		
		int rootIndexInInorder = -1;
		for (int i = inStart; i <= inEnd; i++) {
			if (inorder[i] == rootValue) {
				rootIndexInInorder = i;
				break;
			}
		}
		
		TreeNode rootNode = new TreeNode(rootValue);
		
		int leftCount = rootIndexInInorder - inStart;
		if (leftCount > 0) {
			rootNode.left = buildTree(preorder, start + 1, start + leftCount, inorder, inStart, rootIndexInInorder - 1);
		}
		int rightCount = inEnd - rootIndexInInorder;
		if (rightCount > 0) {
			rootNode.right = buildTree(preorder, start + leftCount + 1, end, inorder, rootIndexInInorder + 1, inEnd);
		}
		
		return rootNode;
	}

}