剑指offer-04：重建二叉树

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

public class Solution04 {


    public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
        TreeNode root = reConstructBinaryTree(pre, 0, pre.length - 1, in, 0, in.length - 1);
        return root;
    }

    //前序遍历{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}
    private TreeNode reConstructBinaryTree(int[] pre, int startPre, int endPre, int[] in, int startIn, int endIn) {

        if (startPre > endPre || startIn > endIn)
            return null;
        TreeNode root = new TreeNode(pre[startPre]);

        for (int i = startIn; i <= endIn; i++)
            if (in[i] == pre[startPre]) {
                root.left = reConstructBinaryTree(pre, startPre + 1, startPre + i - startIn, in, startIn, i - 1);
                root.right = reConstructBinaryTree(pre, i - startIn + startPre + 1, endPre, in, i + 1, endIn);
            }

        return root;

    }

    public static void main(String[] args) {

        int[] pre={1,2,4,7,3,5,6,8};
        int[] in={4,7,2,1,5,3,8,6};
        TreeNode tree=new Solution04().reConstructBinaryTree(pre,in);
        new Solution04().print(tree,tree.val,0);

    }

    private void print(TreeNode tree, int key, int direction) {
        if(tree != null) {
            if(direction==0)    // tree是根节点
                System.out.printf("%2d is root\n", tree.val, key);
            else                // tree是分支节点
                System.out.printf("%2d is %2d's %6s child\n", tree.val, key, direction==1?"right" : "left");

            print(tree.left, tree.val, -1);
            print(tree.right,tree.val,  1);
        }
    }


}

打印结果：

 1 is root
 2 is  1's   left child
 4 is  2's   left child
 7 is  4's  right child
 3 is  1's  right child
 5 is  3's   left child
 6 is  3's  right child
 8 is  6's   left child