Leetcode: Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2024-01-17 14:50:43 发布

chenhaoliu

最新推荐文章于 2024-01-17 14:50:43 发布

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本文链接：https://blog.youkuaiyun.com/chenhaoliu/article/details/8497813

本文介绍了一种使用前序遍历和中序遍历构建二叉树的方法。通过递归方式，利用映射表快速定位根节点在中序遍历中的位置，从而划分左右子树并构造整个二叉树。

/**
 * 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 *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(preorder.size()==0||preorder.size()!=inorder.size())
            return NULL;
        map<int,int> mapper;
        for(int i=0;i<inorder.size();++i)
            mapper[inorder[i]]=i;
        return helper(preorder,inorder,mapper,0,0,preorder.size()-1);
    }
    TreeNode* helper(vector<int>& preorder,vector<int>& inorder,map<int,int>& mapper,int root,int start,int end){
        if(start>end)
            return NULL;
        int r=preorder[root];
        int index=mapper[r];
        TreeNode* node= new TreeNode(r);
        node->left=helper(preorder,inorder,mapper,root+1,start,index-1);
        node->right=helper(preorder,inorder,mapper,root+index-start+1,index+1,end);
        return node;
    }
};