【力扣】105.从前序与中序遍历序列构造二叉树

AC截图

题目

思路

通过前序遍历和中序遍历，可以确定根结点、左子树和右子树。如果我们确定左子树和右子树的范围，就可以递归构建二叉树。

代码

/**
 * 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> map;
    TreeNode* build(vector<int>& preorder,vector<int>& inorder,int preLeft,int preRight,int inLeft,int inRight){
        if(preLeft>preRight){
            return NULL;
        }
        
        int inRoot = map[preorder[preLeft]];
        TreeNode* root = new TreeNode(preorder[preLeft]);
        int leftSize = inRoot-inLeft;
        root->left = build(preorder,inorder,preLeft+1,preLeft+leftSize,inLeft,inRoot-1);
        root->right = build(preorder,inorder,preLeft+leftSize+1,preRight,inRoot+1,inRight);
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        for(int i=0;i<inorder.size();i++){
            map[inorder[i]]=i;
        }
        return build(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
    }
};