LeetCode | Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2025-12-19 16:34:09 发布

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本文介绍了一种通过前序遍历和中序遍历构建二叉树的方法。利用递归方式，找到根节点，并划分左右子树，最终完成整棵树的构建。

题目：

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

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

思路：

与 http://blog.youkuaiyun.com/lanxu_yy/article/details/17383007类似，但是过程相反。

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:
    vector<int> preorder;
    vector<int> inorder;
    int cur;
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if(preorder.size() == 0)
            return NULL;
        else
        {
            cur = 0;
            this->preorder = preorder;
            this->inorder = inorder;
            return buildSubTree(0, inorder.size()-1);
        }
    }
    
    TreeNode* buildSubTree(int from, int to)
    {
        TreeNode * root = new TreeNode(preorder[cur]);
        int index = findValInInOrder(preorder[cur]);
        cur++;
        if(index <= from)
        {
            root->left = NULL;
        }
        else
        {
            root->left = buildSubTree(from, index-1);
        }
        if(index >= to)
        {
            root->right = NULL;
        }
        else
        {
            root->right = buildSubTree(index+1,to);
        }
        return root;
    }
    
    int findValInInOrder(int val)
    {
        for(int i=0;i<inorder.size();i++)
        {
            if(val == inorder[i])
            {
                return i;
            }
        }
        return -1;
    }
};