LeetCode——Construct Binary Tree from Inorder and Postorder Traversal

pickless

于 2013-09-22 00:19:46 发布

阅读量860

点赞数

CC 4.0 BY-SA版权

分类专栏： Online Judge 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/pickless/article/details/11884033

Online Judge 专栏收录该内容

134 篇文章

订阅专栏

本文详细介绍了如何通过给定的中序和后序遍历序列来构建二叉树，包括算法实现及注意事项。

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.

» Solve this problem

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    TreeNode *build(vector<int> &inorder, vector<int> &postorder, int il, int ir, int pl, int pr) {
        TreeNode *root;
        if (il > ir || pl > pr) {
            root = NULL;            
        }
        else {
            root = new TreeNode(postorder[pr]);
            int i;
            for (i = il; i <= ir && inorder[i] != postorder[pr]; i++);
            root->left = build(inorder, postorder, il, i - 1, pl, pl + i - il - 1);
            root->right = build(inorder, postorder, i + 1, ir, pl + i - il, pr - 1);
        }
        return root;
    }

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, postorder, 0, inorder.size() - 1, 0, postorder.size() - 1);
    }
};