[LeetCode]105. Construct Binary Tree from Preorder and Inorder Traversal

1. Construct Binary Tree from Preorder and Inorder Traversal
   Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

---

已经假定了二叉树中不存在重复的值！

/** * 比较简洁，但反复复制vector，浪费空间 *//** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        if(preorder.empty() || inorder.empty())            return NULL;        TreeNode* root = new TreeNode(preorder[0]);        int i = 0;        for(; i < inorder.size(); ++i){            if(inorder[i] == root->val)                break;        }        // root如果有左子树,长度为i        if(i > 0){            vector<int> pre_left(preorder.begin()+1, preorder.begin()+1+i);            vector<int> in_left(inorder.begin(), inorder.begin()+i);            root->left = buildTree(pre_left, in_left);        }        // root如果有右子树        if(i < inorder.size() - 1){            vector<int> pre_right(preorder.begin()+1+i, preorder.end());            vector<int> in_right(inorder.begin()+i+1, inorder.end());            root->right = buildTree(pre_right, in_right);        }        return root;    }};

/** * 直接传入索引，不用开辟新的vector，节省空间 */ /** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        if(preorder.empty() || inorder.empty())            return NULL;        return build(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);    }    // preorder[pre_left, pre_right] 左闭右闭区间    TreeNode* build(vector<int>& preorder, int pre_left, int pre_right, vector<int>& inorder, int in_left, int in_right){        TreeNode* root = new TreeNode(preorder[pre_left]);        // 找到root在inorder中的位置索引        int i = in_left;        for(; i <= in_right; ++i){            if(inorder[i] == root->val)                break;        }        // 在这里已经保证了两个局部使用的vector不会为空，所以最开始不需要检查索引        // 如果存在左子树,长度为 i-in_left        if(i > in_left){            root->left = build(preorder, pre_left+1, pre_left+i-in_left, inorder, in_left, i-1);        }        // 如果存在右子树        if(i < in_right){            root->right = build(preorder, pre_left+i-in_left+1, pre_right, inorder, i+1, in_right);        }        return root;    }};