劍指Offer P62 面試題7：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 Solution 
{
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) 
    {
        if (pre == null || in == null || pre.length == 0 || in.length == 0)
            return null;
        TreeNode root = new TreeNode(pre[0]);
        for (int location = 0; location < in.length; location++)
        {
            if (pre[0] == in[location])
            {
                // 左子樹
                int[] leftPre = new int[location];
                int[] leftIn = new int[location];
                for (int i = 0; i < location; i++)
                {
                    leftPre[i] = pre[i + 1];
                    leftIn[i] = in[i];
                }
                root.left = reConstructBinaryTree(leftPre, leftIn);                    
                
                // 右子樹
                int[] rightPre = new int[in.length - 1 - location];
                int[] rightIn = new int[in.length - 1 - location];
                for (int i = 0; i < in.length - 1 - location; i++)
                {
                    rightPre[i] = pre[location + 1 + i];
                    rightIn[i] = in[location + 1 + i];
                }
                root.right = reConstructBinaryTree(rightPre, rightIn);                    
            }
        }
        return root;
    }
}
 

遞迴呼叫的時候一定要寫

root.left  = 

root.right = 

如果允許匯入相關包的話，就不用建立新陣列了

root.left = reConstructBinaryTree(Arrays.copyOfRange(pre, 1, location+1), Arrays.copyOfRange(in, 0, location));
root.right = reConstructBinaryTree(Arrays.copyOfRange(pre, location+1, pre.length), Arrays.copyOfRange(in, location+1, in.length));