LeetCode: Construct Binary Tree from Inorder and Postorder Traversal

构建二叉树：从后序遍历和中序遍历出发

最新推荐文章于 2022-01-01 11:35:53 发布

chenhaoliu

最新推荐文章于 2022-01-01 11:35:53 发布

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本文链接：https://blog.youkuaiyun.com/chenhaoliu/article/details/8497966

本文介绍如何利用后序遍历和中序遍历的结果来构建一棵二叉树，详细解释了构建过程中的关键步骤和实现代码。

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