Rebuild binary tree

重建二叉树

输入一棵二叉树前序遍历和中序遍历的结果,请重建该二叉树.

注意 ：

• 二叉树中每个节点的值都互不相同;
• 输入的前序遍历和中序遍历一定合法;

数据范围

树中节点数量范围$[0,100]$.

样例

给定：
前序遍历是：[3, 9, 20, 15, 7]
中序遍历是：[9, 3, 15, 20, 7]

返回：[3, 9, 20, null, null, 15, 7, null, null, null, null]
返回的二叉树如下所示：
    3
   / \
  9  20
    /  \
   15   7

算法

（递归)$O(n)$

递归建立整棵二叉树：先递归创建左右子树,然后创建根节点,并让指针指向两棵子树.

具体步骤如下：

1. 先利用前序遍历找根节点：前序遍历的第一个数,就是根节点的值;
2. Find the position of the current root node in an in-order traversal $k$ ,则 $k$ 左边是左子树的中序遍历``,右边是右子树的中序遍历;
3. 假设左子树的中序遍历的长度是 $l$ ,则在前序遍历中,根节点后面的 $l$ 个数,是左子树的前序遍历,剩下的数是右子树的前序遍历;
4. 有了左右子树的前序遍历和中序遍历,我们可以先递归创建出左右子树,然后再创建根节点;

时间复杂度分析

初始化时,用哈希表（unordered_map<int, int>）Record the position of each value in the in-order traversal,这样我们在递归到每个节点时,An operation that finds the location of the root node in an inorder traversal,只需要 $O(1)$ 的时间.此时,创建每个节点需要的时间是 $O(1)$ ,所以总时间复杂度是 $O(n)$.

变量说明

• pos：Record the position of each node in the in-order traversal sequence.
• pre： topic input data,Represents the preorder traversal sequence of the current binary tree.
• in： topic input data,Represents the inorder traversal sequence of the current binary tree.
• k： Indicates the position of the current root node in the given inorder traversal sequence(分割左右子树).
• pl：Indicates the starting position of the corresponding range in the preorder traversal sequence of the tree represented by the current root node.
• pr：Indicates the end position of the corresponding range in the preorder traversal sequence of the tree represented by the current root node.
• il：Indicates the starting position of the corresponding range in the in-order traversal sequence of the tree represented by the current root node.
• ir：Indicates the end position of the corresponding range in the in-order traversal sequence of the tree represented by the current root node.

/**
 * 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:
    // Record the position of each point in the inorder traversal sequence
    unordered_map<int, int> pos;
    
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int n = preorder.size();
        for (int i = 0; i < n; i ++ )
             pos[inorder[i]] = i;
        return dfs(preorder, inorder, 0, n - 1, 0, n - 1);
    }
    
    // DFS递归遍历二叉树
    TreeNode* dfs(vector<int>& pre, vector<int>& in, int pl, int pr, int il, int ir) {
        if (pl > pr) return NULL;
        int k = pos[pre[pl]]; // The position of the current root node in the inorder traversal sequence
        TreeNode* root = new TreeNode(pre[pl]);
        root->left = dfs(pre, in, pl + 1, pl + k - il, il, k - 1);
        root->right = dfs(pre, in, pl + k + 1 -il, pr, k + 1, ir);
        return root;
    }
};