006-重建二叉树

#include <iostream>
#include <vector>

using namespace std;

//  调试开关
#define __tmain main

#ifdef __tmain

#define debug cout

#else

#define debug 0 && cout

#endif // __tmain


#ifdef __tmain
struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;

    TreeNode(int x)
    : val(x), left(NULL), right(NULL)
    {
    }

    //
    static void PreOrder(TreeNode *root)
    {
        if(root == NULL)
        {
            return;
        }
        cout <<root->val;
        PreOrder(root->left);
        PreOrder(root->right);
    }

    static void InOrder(TreeNode *root)
    {
        if(root == NULL)
        {
            return;
        }
        InOrder(root->left);
        cout <<root->val;
        InOrder(root->right);
    }

 };
#endif // __tmain

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution
{
public:
    struct TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in)
    {
        //  前序遍历的长度跟中序遍历的长度应该相同
        if(pre.size( ) != in.size( ))
        {
            debug <<"the length of PRE and IN should be smae" <<endl;
            return NULL;
        }

        // 长度不能为0
        int size = pre.size( );
        if(size == 0)
        {
            debug <<"it's a NULL tree(length = 0)" <<endl;
            return NULL;
        }

        int length = pre.size( );
        debug <<"the length of your tree = " <<length <<endl;
        int value = pre[0];      //  前序遍历的第一个结点是根节点
        TreeNode *root = new TreeNode(value);

        debug <<"the root is" <<root->val <<endl;
        //  在中序遍历中查找到根的位置
        int rootIndex = 0;
        for(rootIndex = 0; rootIndex < length; rootIndex++)
        {
            if(in[rootIndex] == value)
            {
                debug <<"find the root at " <<rootIndex <<" in IN" <<endl;
                break;
            }
        }
        if(rootIndex >= length)
        {
            debug <<"can't find root (value = " <<value <<") in IN" <<endl;
            return NULL;
        }

        ///  区分左子树和右子树
        ///  中序遍历中, 根左边的就是左子数, 右边的就是右子树
        ///  前序遍历中, 根后面是先遍历左子树, 然后是右子树

        ///  首先确定左右子树的长度, 从中序遍历in中确定
        int leftLength = rootIndex;
        int rightLength = length - 1 - rootIndex;
        debug <<"left length = " <<leftLength <<", rightLength = " <<rightLength <<endl;
        vector<int> preLeft(leftLength), inLeft(leftLength);
        vector<int> preRight(rightLength), inRight(rightLength);
        for(int i = 0; i < length; i++)
        {
            if(i < rootIndex)
            {
                //  前序遍历的第一个是根节点, 根后面的(leftLegnth = rootIndex) - 1个节点是左子树, 因此是i+1
                preLeft[i] = pre[i + 1];
                //  中序遍历前(leftLength = rootIndex) - 1个节点是左子树, 第rootIndex个节点是根
                inLeft[i] = in[i];
                debug <<preLeft[i] <<inLeft[i] <<" ";

            }
            else if(i > rootIndex)
            {
                //  前序遍历的第一个是根节点, 根后面的(leftLegnth = rootIndex) - 1个节点是左子树, 后面是右子树
                preRight[i - rootIndex - 1] = pre[i];
                //  中序遍历前(leftLength = rootIndex) - 1个节点是左子树, 第rootIndex个节点是根, 然后是右子树
                inRight[i - rootIndex - 1] = in[i];
                debug <<preRight[i - rootIndex - 1] <<inRight[i - rootIndex - 1] <<" ";

            }
        }
        debug <<endl <<"the left tree" <<endl;
        for(int i = 0; i < leftLength; i++)
        {
            debug <<preLeft[i] <<inLeft[i] <<" ";
        }
        debug <<endl;
        debug <<"the right tree" <<endl;
        for(int i = 0; i < rightLength; i++)
        {
            debug <<preRight[i] <<inRight[i] <<" ";
        }
        debug <<endl;


        root->left = reConstructBinaryTree(preLeft, inLeft);
        root->right = reConstructBinaryTree(preRight, inRight);

        return root;
    }





};

int __tmain( )
{
    int pre[] = { 1, 2, 4, 7, 3, 5, 6, 8 };
    int in[] = { 4, 7, 2, 1, 5, 3, 8, 6 };

    vector<int> preOrder(pre, pre + 8);
    vector<int>  inOrder( in,  in + 8);

    Solution solu;
    TreeNode *root = solu.reConstructBinaryTree(preOrder, inOrder);

    cout <<"PreOrder";
    TreeNode::PreOrder(root);
    cout <<endl;

    cout <<"InOrder ";
    TreeNode::InOrder(root);
    cout <<endl;

    return 0;
}