LeetCode-剑指7-重建二叉树

题目同LeetCode-105。

1、递归

利用哈希表记录前序遍历中每个元素在中序遍历中的位置用于查询。

class Solution {
public:
    unordered_map<int, int> preToin;

    TreeNode *build(const vector<int> &preorder, const vector<int> &inorder, int preorder_low, int preorder_high, int inorder_low, int inorder_high) {
        if (preorder_low > preorder_high)
            return nullptr;
        int preorder_root = preorder_low;
        int inorder_root = preToin[preorder[preorder_root]];
        TreeNode *root = new TreeNode(preorder[preorder_root]);
        int leftTreeLength = inorder_root - inorder_low;
        root->left = build(preorder, inorder, preorder_low + 1, preorder_low + leftTreeLength, inorder_low, inorder_root - 1);
        root->right = build(preorder, inorder, preorder_low + leftTreeLength + 1, preorder_high, inorder_root + 1, inorder_high);
        return root;
    }

    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if (preorder.size() == 0) return nullptr;
        for (int i = 0; i < inorder.size(); ++i) {
            preToin[inorder[i]] = i;
        }
        return build(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);
    }
};

2、迭代

利用栈暂存构建二叉树时左子树中的值。

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;
    }
};