剑指offer | 重建二叉树 | 25

目录

• 思路分析

思路分析

首先需要注意: 本题是没有重复值的!
优化: 利用哈希表,快速找到中序值对应的下标值

cpp

/**
 * 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:

    unordered_map<int,int> d; // 优化: 记录中序序列的位置
    vector<int> preorder,inorder; // 为了dfs方便,两个序列定义为全局
    // 先序的序列范围 和 中序的序列范围 (闭区间)
    TreeNode* dfs(int pl,int pr,int il,int ir){
        if(pl>pr)return NULL;
        auto root = new TreeNode(preorder[pl]); // 根节点的值
        int k = d[root->val];
        auto left = dfs(pl+1,pl+k-il,il,k-1);
        auto right = dfs(pl+k-il+1,pr,k+1,ir);
        root->left = left,root->right=right;
        return root;
    }
    
    TreeNode* buildTree(vector<int>& _preorder, vector<int>& _inorder) {
        preorder = _preorder,inorder=_inorder;
        for(int i=0;i<inorder.size();i++)d[inorder[i]]=i;
        
        return dfs(0,preorder.size()-1,0,inorder.size()-1);
        
    }
};

python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution(object):
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        d = dict()
        def dfs(pl,pr,il,ir):
            if pl>pr:return None
            root = TreeNode(preorder[pl])
            k = d[root.val]
            left = dfs(pl+1,pl+k-il,il,k-1)
            right = dfs(pl+k-il+1,pr,k+1,ir)
            root.left,root.right=left,right
            return root
        
        for i,x in enumerate(inorder):
            d[x]=i
        
        return dfs(0,len(preorder)-1,0,len(inorder)-1)