2018.7.8学习笔记

剑指Offer P62 面试题7：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 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])
            {
                // 左子树
                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);                    
                
                // 右子树
                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;
    }
}

递归调用的时候一定要写

root.left  = 

root.right = 

如果允许导入相关包的话，就不用建立新数组了

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));