輸入某二叉樹的前序遍歷和中序遍歷的結果，請重建出該二叉樹。假設輸入的前序遍歷和中序遍歷的結果中都不含重複的數字。例如輸入前序遍歷序列{1,2,4,7,3,5,6,8}和中序遍歷序列{4,7,2,1,5,3,8,6}，則重建二叉樹並返回。

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;
    }

    //前序遍歷{1,2,4,7,3,5,6,8}和中序遍歷序列{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是根節點
                System.out.printf("%2d is root\n", tree.val, key);
            else                // tree是分支節點
                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);
        }
    }


}

列印結果：

 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