题目描述


法一）递归

复杂度分析

代码如下

class Solution {
private:
    unordered_map<int, int> index;

public:
    TreeNode* myBuildTree(const vector<int>& preorder, const vector<int>& inorder, int preorder_left, int preorder_right, int inorder_left, int inorder_right){
        if(preorder_left > preorder_right){
            return nullptr;
        }
        int preorder_root = preorder_left;
        int inorder_root = index[preorder[preorder_root]];

        TreeNode* root = new TreeNode(preorder[preorder_root]);
        int size_left_subtree = inorder_root - inorder_left;
        root->left = myBuildTree(preorder, inorder, preorder_left+1, preorder_left+size_left_subtree, inorder_left, inorder_root-1);
        root->right = myBuildTree(preorder, inorder, preorder_left+size_left_subtree+1, preorder_right, inorder_root+1, inorder_right);
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int n = preorder.size();
        for(int i=0; i<n; ++i){
            index[inorder[i]]  = i;
        }
        return myBuildTree(preorder, inorder, 0, n-1, 0, n-1);
    }
};

法二）迭代




复杂度分析

代码如下

class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if (!preorder.size()) {
            return nullptr;
        }
        TreeNode* root = new TreeNode(preorder[0]);
        stack<TreeNode*> stk;
        stk.push(root);
        int inorderIndex = 0;
        for (int i = 1; i < preorder.size(); ++i) {
            int preorderVal = preorder[i];
            TreeNode* node = stk.top();
            if (node->val != inorder[inorderIndex]) {
                node->left = new TreeNode(preorderVal);
                stk.push(node->left);
            }
            else {
                while (!stk.empty() && stk.top()->val == inorder[inorderIndex]) {
                    node = stk.top();
                    stk.pop();
                    ++inorderIndex;
                }
                node->right = new TreeNode(preorderVal);
                stk.push(node->right);
            }
        }
        return root;
    }
};