105. 从前序与中序遍历序列构造二叉树

中等 · 递归

题目

105. 从前序与中序遍历序列构造二叉树 官方题解

方法一: 递归

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* build(vector<int>& preorder, vector<int>& inorder, int l1, int r1, int l2, int r2) {
        if (l1 > r1) return nullptr;

        int len;
        for (int i = l2; i <= r2; i ++ ) {
            if (inorder[i] == preorder[l1]) {
                len = i - l2;
            }
        }

        TreeNode* root = new TreeNode(preorder[l1]);
        root->left = build(preorder, inorder, l1 + 1, l1 + len, l2, l2 + len - 1);
        root->right = build(preorder, inorder, l1 + len + 1, r1, l2 + len + 1, r2);

        return root;
    }

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return build(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);
    }
};

时间复杂度:O(n^2)

空间复杂度:O(n)

##方法二:递归 + 哈希表

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    unordered_map<int, int> umap;

    TreeNode* build(vector<int>& preorder, vector<int>& inorder, int l1, int r1, int l2, int r2) {
        if (l1 > r1) return nullptr;

        int len = umap[preorder[l1]] - l2;

        TreeNode* root = new TreeNode(preorder[l1]);
        root->left = build(preorder, inorder, l1 + 1, l1 + len, l2, l2 + len - 1);
        root->right = build(preorder, inorder, l1 + len + 1, r1, l2 + len + 1, r2);

        return root;
    }

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

时间复杂度:O(n)

空间复杂度:O(n)