leetcode-105. 从前序与中序遍历序列构造二叉树

 

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    unordered_map<int,int> in_map;
    vector<int> pre_order;
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(preorder.size()==0)
            return NULL;
        cout<<"flag"<<endl;
        for(int i = 0; i < inorder.size(); i++){
            in_map[inorder[i]] = i;
            pre_order.push_back(preorder[i]);
        }
        return recur(0,0,inorder.size()-1);
    }
    // pre:在先序遍历中的跟节点的下标。
    // left:在中序遍历中的左边界
    // right:在中序遍历中的右边界
    TreeNode* recur(int pre,int left,int right){
        if(left>right)
            return NULL;

        int mid = in_map[pre_order[pre]];
        // cout<<"mid"<<mid<<endl;
        // cout<<"pre_order[pre]:"<<pre_order[pre]<<endl;
        TreeNode* root = new TreeNode(pre_order[pre]);
        root->left = recur(pre+1,left,mid-1);
        root->right = recur(pre+(mid-left)+1,mid+1,right);
        return root;
    }
};