题解 | #重建二叉树#

/**
 * 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) {
        int s1=0,e1 = pre.size()-1,s2=0,e2 = vin.size()-1;
        if(e1!=e2) {
            return nullptr;
        }
        map<int,int> mark = map<int, int>();
        for(int i=0;i<vin.size();i++) {
            mark[vin[i]] = i;
        }
        return help(pre,s1,e1,vin,s2,e2,mark);
    }
    
    TreeNode* help(vector<int> pre,int s1,int e1,vector<int> vin,int s2,int e2,map<int,int>mark) {
        if(e1-s1!=e2-s2 || s1>s1 || s2>e2) {
            return nullptr;
        }
        TreeNode* root = new TreeNode(pre[s1]);
        int k = mark[pre[s1]];
        TreeNode* left = help(pre,s1+1,s1+k-s2,vin,s2,k-1,mark);
        TreeNode* right = help(pre,s1+k-s2+1,e1,vin,k+1,e2,mark);
        root->left = left;
        root->right = right;
        return root;
    }
};