105. Construct Binary Tree from Preorder and Inorder Traversal

问题描述：

Given preorder and inorder traversal of a tree, construct the binary tree.

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

解题思路：

我学习了大佬的解法

用辅助函数来辅助简历，根据先序遍历的特点，数组第一个点一定是根节点，而中序遍历中，根节点放在中间。

我们可以提取出根节点后，将数组分为左子树和右子树

对于中序遍历我们可以找到与根节点值相同的点并得到它的坐标 i ，那么左子树和右子树的范围分别为：

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　　左子树：[iL, i-1] 个数： i - iL 

　　右子树:  [i+1, iR]  个数： iR - i

我们可以推出前序遍历中左右子树的位置：

　　左子树：[pL+1, pL+ i - iL]

　　右子树：[pL + i - iL + 1, pR]

代码：

/**
 * 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) {
        return build(preorder, 0, preorder.size()-1, inorder, 0, inorder.size() -1 );
    }
private:
    TreeNode* build(vector<int>& preorder, int pL, int pR, vector<int>& inorder, int iL, int iR){
        if(pL > pR || iL > iR){
            return NULL;
        }
        int i = 0;
        for(i = iL; i<= iR; i++){
            if(inorder[i] == preorder[pL])
                break;
        }
        TreeNode* cur = new TreeNode(preorder[pL]);
        cur->left = build(preorder, pL+1, pL + i - iL, inorder, iL, i-1);
        cur->right = build(preorder, pL + i-iL+1, pR, inorder, i+1, iR);
        return cur;
    }
};

该解法总耗时35ms

来看一下10ms的解法：看不懂：）

/**
 * 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) {
        int p = 0, i = 0;
        int end = inorder.size();
        return dfs(preorder, inorder, p, i, end);
    }
    
    TreeNode* dfs(vector<int>& preorder, vector<int>& inorder, int& p, int& i, int end)
    {
        if(i < inorder.size() && (end == inorder.size() || inorder[i] != preorder[end]))
        {
            TreeNode* ret = new TreeNode(preorder[p++]);
            ret->left = dfs(preorder, inorder, p, i, p-1);
            ++i;
            ret->right = dfs(preorder, inorder, p, i, end);
            return ret;
        }
        else return NULL;
    }
};