Leetcode Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2025-08-12 12:49:47 发布

rosepicker

最新推荐文章于 2025-08-12 12:49:47 发布

阅读量220

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/rosepicker/article/details/73930188

leetcode 专栏收录该内容

124 篇文章

订阅专栏

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。通过递归算法，将遍历序列转化为具体的二叉树结构。该方法假设树中不存在重复元素。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

代码如下：

/**
 * Definition for a binary tree node.
 * 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) {
        int plen = preorder.size(),ilen = inorder.size();
        
        return buildTree(preorder,0,plen-1,inorder,0,ilen-1);
    }
    
    TreeNode* buildTree(vector<int>& preorder,int pstart,int pend,vector<int>& inorder,int istart,int iend)
    {
        if(pstart > pend || istart > iend)
            return NULL;
            
        TreeNode* root=new TreeNode(preorder[pstart]);
        
        int i;
        for(i=0;i<=iend-istart+1;i++)
            if(preorder[pstart] == inorder[istart+i])
                break;
        root->left = buildTree(preorder,pstart+1,pstart+i,inorder,istart,istart+i-1);
        root->right = buildTree(preorder,pstart+i+1,pend,inorder,istart+i+1,iend);
        return root;
    }
};