剑指 Offer 07. 重建二叉树（中序加前序）

题目描述

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    
    
public:
    map<int,int> index;
    TreeNode* helper(vector<int>& preorder,int p_start,int p_end,vector<int>& inorder,int i_start,int i_end){
    
    
        //判断是否为空
        if(p_end < p_start)
           return NULL;
        //建立根节点
        TreeNode* root = new TreeNode(preorder[p_start]);
        //中序遍历中根节点的位置
        int inorder_root = index[preorder[p_start]];
        //左子树中节点的个数
        int left_num = inorder_root - i_start;

        root->left =  helper(preorder,p_start+1,p_start+left_num,inorder,i_start,inorder_root-1);
        root->right = helper(preorder,p_start+left_num+1,p_end,inorder,inorder_root+1,i_end);
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
    
    
        int len = inorder.size();
        //之后不用遍历中序遍历数组
        for(int i = 0;i<len;i++){
    
    
            index[inorder[i]] = i;
        } 
        TreeNode* root = helper(preorder,0,len-1,inorder,0,len-1);
        return root;
    }
};