牛客_剑指offer_重建二叉树，再后续遍历_递归思想_分两端

   总结：

   重建二叉树：其实就是根据前序和中序重建得到二叉树，得到后续，只要输出那边设置输出顺序即可

[编程题]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。

完整通过代码：

先新建一个二叉树的类

public class TreeNode {
    int val;
    TreeNode left;
   TreeNode right;
    TreeNode(int x) { val = x; }
}

　再　

public class reConstructBinaryTree {
     public static TreeNode reConstructBinaryTree(int [] pre,int [] in) {
               if(pre==null||in==null){       return null;            }
               TreeNode tree = reConstructCore(pre,in,0,pre.length-1,0,in.length-1);
               return tree;
     }
     /**
        *核心算法，preStart和preEnd是起始下标和结束下标
        **/
      public static TreeNode reConstructCore(int[] pre,int[] in,int preStart,int preEnd,int inStart,int inEnd){
            TreeNode tree = new TreeNode(pre[preStart]);
            tree.left = null;
            tree.right= null;
            if(preStart==preEnd&&inStart==inEnd){ return tree;   }
            //记录中序遍历中等于前序遍历的第一位的下标
            int inCenter = 0;
            for(inCenter = inStart;inCenter<inEnd;inCenter++){
                if(in[inCenter]==pre[preStart]){
                    break;
                }
            }
            //左子树的长度
            int leftTreeLength = inCenter-inStart;
            //右子数的长度
            int rightTreeLength = inEnd-inCenter;
            if(leftTreeLength>0){
               tree.left = reConstructCore(pre,in,preStart+1,preStart+leftTreeLength,inStart,inCenter-1);
            }
            if(rightTreeLength>0){
               tree.right = reConstructCore(pre,in,preStart+leftTreeLength+1,preEnd,inCenter+1,inEnd);
            }
          return tree;   

      }
      public static void traverseBinTreeRDL(TreeNode node){
          if (node==null) {    return;            }
          if (node.left!=null) {
              traverseBinTreeRDL(node.left);    }
          if(node.right!=null) {
              traverseBinTreeRDL(node.right);   }
          System.out.println(node.val);
        }
    public static void main(String[] arg0){
        int pre[]= {1,2,4,7,3,5,6,8};
        int in[] = {4,7,2,1,5,3,8,6};
        TreeNode tree = reConstructBinaryTree(pre, in);
        traverseBinTreeRDL(tree);
     }

}

输出：

7
4
2
5
8
6
3
1