重建二叉树

题目描述

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。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        if(pre.empty() || vin.empty())
            return NULL;
        return constructTree(pre, 0, pre.size() - 1, vin, 0, vin.size() - 1);
    }
    
    TreeNode* constructTree(vector<int> pre, int preStart, int preEnd, vector<int> vin, int vinStart, int vinEnd)
    {
        int rootValue = pre[preStart];
        TreeNode* root = new TreeNode(rootValue);
        
        if(preStart == preEnd)
        {
            if(vinStart == vinEnd && pre[preStart] == vin[vinStart])
                return root;
          
        }
        
        
        //中序遍历查找根节点
        //int rootInorder = vin[vinStart];
        int tmpStart = vinStart;
        while(tmpStart <= vinEnd && vin[tmpStart] != rootValue)
            tmpStart++;
        
        int leftLength = tmpStart - vinStart;
        int leftPreorderEnd = preStart + leftLength;
        
        if(leftLength > 0)
        {
            //构建左子树
            root->left = constructTree(pre, preStart + 1, leftPreorderEnd, vin, vinStart, tmpStart - 1);
        }
        if(leftLength < vinEnd - vinStart)
        {
            root->right = constructTree(pre, leftPreorderEnd + 1, preEnd, vin, tmpStart + 1, vinEnd);
        }
        return root;
        
    }
};