LeetCode 105. 从前序与中序遍历序列构造二叉树

题目描述

105. 从前序与中序遍历序列构造二叉树

解法：

具体从前序结果和中序结果构造二叉树的流程就不再介绍，我们这里给出一幅图说一下关于索引的确定

int leftSize = index - inStart;

root->left = build(preorder, preStart + 1, preStart + leftSize,
                  inorder, inStart, index - 1);

root->right = build(preorder, preStart + leftSize + 1, preEnd,
                   inorder, index + 1, inEnd);

完整的递归构造代码如下所示

/** * 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* buildTree(vector<int>& preorder, vector<int>& inorder) {
    
        return helper(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
    }
    
    TreeNode* helper(vector<int>& preorder, int prestart, int preend, vector<int>& inorder, int instart, int inend){
    
        if (prestart > preend) return nullptr;

        int rootval = preorder[prestart], indx = 0;
        for (int i = instart; i <= inend; i++)
            if (inorder[i] == rootval) indx = i;

        int leftsize = indx - instart;
        TreeNode* root = new TreeNode(rootval);
        root->left = helper(preorder, prestart + 1, prestart + leftsize, inorder, instart, indx - 1);
        root->right = helper(preorder, prestart + leftsize + 1, preend, inorder, indx + 1, inend);
        return root;
    }
};