LeetCode hot100-105从前序与中序遍历序列构造二叉树

啊哦呃咦唔鱼2026-04-10 21:08

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class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(preorder.empty()||inorder.empty())  return nullptr;
        vector<int> lpre,lin,rpre,rin;
        TreeNode* ans=new TreeNode(preorder[0]);;
        int flg=0;
          // 分割中序：左子树和右子树（不含根节点）
        for(int i=0;i<inorder.size();++i){
           if(inorder[i]==preorder[0]){
            flg=i;
            break;
           }
        }
        for(int i=0;i<flg;++i){
            lin.push_back(inorder[i]);
        }
        for(int i=flg+1;i<inorder.size();++i){
            rin.push_back(inorder[i]);
        }
        // 分割前序：根据左子树的大小
        for(int i=1;i<=lin.size();++i){
                lpre.push_back(preorder[i]);
        }

        for(int i = 1 + lin.size(); i < preorder.size(); ++i) {
            rpre.push_back(preorder[i]);
        }
      
        ans->left=buildTree(lpre,lin);
        ans->right=buildTree(rpre,rin);

        return ans;
    }
};

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class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if (preorder.empty() || inorder.empty()) return nullptr;
        
        // 根节点是前序遍历的第一个元素
        int rootVal = preorder[0];
        TreeNode* root = new TreeNode(rootVal);
        
        // 在中序遍历中找到根节点位置
        int rootIndex = 0;
        for (int i = 0; i < inorder.size(); ++i) {
            if (inorder[i] == rootVal) {
                rootIndex = i;
                break;
            }
        }
        
        // 分割中序遍历
        vector<int> leftInorder(inorder.begin(), inorder.begin() + rootIndex);
        vector<int> rightInorder(inorder.begin() + rootIndex + 1, inorder.end());
        
        // 分割前序遍历
        vector<int> leftPreorder(preorder.begin() + 1, 
                                 preorder.begin() + 1 + leftInorder.size());
        vector<int> rightPreorder(preorder.begin() + 1 + leftInorder.size(), 
                                  preorder.end());
        
        // 递归构建左右子树
        root->left = buildTree(leftPreorder, leftInorder);
        root->right = buildTree(rightPreorder, rightInorder);
        
        return root;
    }
};

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class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return build(preorder, 0, preorder.size() - 1,
                     inorder, 0, inorder.size() - 1);
    }
    
private:
    TreeNode* build(vector<int>& preorder, int preStart, int preEnd,
                    vector<int>& inorder, int inStart, int inEnd) {
        if (preStart > preEnd || inStart > inEnd) return nullptr;
        
        TreeNode* root = new TreeNode(preorder[preStart]);
        
        // 在中序中找到根节点
        int rootIndex = inStart;
        while (inorder[rootIndex] != preorder[preStart]) rootIndex++;
        
        int leftSize = rootIndex - inStart;
        
        // 递归构建
        root->left = build(preorder, preStart + 1, preStart + leftSize,
                          inorder, inStart, rootIndex - 1);
        root->right = build(preorder, preStart + leftSize + 1, preEnd,
                           inorder, rootIndex + 1, inEnd);
        
        return root;
    }
};