标签：malloc   return   vector   index   turn   遍历   ret   dex   pre   

剑指OFFER 重建二叉树

使用的是递归的算法,不断基于前序遍历的父亲结点把中序遍历数组分成两半.

其状态图,之后再补充

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void reconstruct_recur(TreeNode* &node,vector<int> &pre,vector<int> &vin,int pre_left,int pre_right,int vin_left,int vin_right)
    {
        if(pre_left>pre_right || vin_left>vin_right)return;
        node =(TreeNode*)malloc(sizeof(TreeNode));
        node->val = pre[pre_left];
        node->left = NULL;
        node->right = NULL;
        auto vin_it = find(vin.begin()+vin_left,vin.end()+vin_right,pre[pre_left]);
        int vin_index = vin_it - vin.begin() - vin_left;//代表着vin从"中间"划分后左边元素的个数
        reconstruct_recur(node->left,pre,vin,pre_left+1,pre_left+vin_index,vin_left,vin_left+vin_index-1);//左边
        reconstruct_recur(node->right,pre,vin,pre_left+vin_index+1,pre_right,vin_left+vin_index+1,vin_right);//右边
    }

    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        TreeNode* root = NULL;
        reconstruct_recur(root,pre,vin,0,pre.size()-1,0,vin.size()-1);
        return root;
    }
};

剑指OFFER 重建二叉树

标签：malloc   return   vector   index   turn   遍历   ret   dex   pre   

原文地址：https://www.cnblogs.com/virgildevil/p/12182426.html