Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2021-02-03 15:56:42 发布

zephyr_be_brave

最新推荐文章于 2021-02-03 15:56:42 发布

阅读量597

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode习题分析

本文链接：https://blog.youkuaiyun.com/zephyr_be_brave/article/details/8907286

leetcode习题分析专栏收录该内容

137 篇文章

订阅专栏

本文介绍了一种通过给定的中序和后序遍历序列来构造二叉树的方法。使用递归方式找到根节点，并根据中序遍历划分左右子树，最终构建完整的二叉树。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

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

/**
 * 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
        return build(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1);  
    }
    
    TreeNode *build(vector<int> &inorder,int s1,int e1, vector<int> &postorder,int s2,int e2)  
    {  
        if(s1 > e1 || s2 > e2) return NULL;  
          
          
        vector<int>::iterator it = find(inorder.begin(),inorder.end(),postorder[e2]);  
        int index = it - inorder.begin();  
         
        TreeNode *root = new TreeNode(postorder[e2]);  
        TreeNode *left = build(inorder,s1,index-1,postorder,s2,s2+index-s1-1);  
        TreeNode *right = build(inorder,index+1,e1,postorder,s2+index-s1,e2-1);  
          
        root->left = left;  
        root->right = right;  
        return root;  
    }  
};

152 milli secs