LeetCode 105. 从前序与中序遍历序列构造二叉树(C++)

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class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        TreeNode* root = build(preorder,inorder,0,preorder.size()-1,0,inorder.size()-1);
        return root;
    }
    TreeNode* build(vector<int>& preorder,vector<int>& inorder,int preL,int preR,int inL,int inR){
        if(preL>preR||inL>inR) return NULL;
        TreeNode* root = new TreeNode(preorder[preL]);
        //root->val = preorder[preL];
        int leftlen = 0;
        for(int i = inL;i<=inR;i++){
            if(inorder[i]==preorder[preL]){
                leftlen = i - inL;
                break;
            }
        }
        root->left = build(preorder,inorder,preL+1,preL+1+leftlen-1,inL,inL+leftlen-1);
        root->right = build(preorder,inorder,preL+leftlen+1,preR,inL+leftlen+1,inR);
        return root;
    }

};

LeetCode 105. 从前序与中序遍历序列构造二叉树(C++)

原文：https://www.cnblogs.com/shuibeng/p/13626914.html