[Leetcode]105. Construct Binary

105. 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.

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

题意：中序加前序构建二叉树
思路，从前序中找到在根结点在中序中的位置，然后找出左右子树，递归

/**
 * 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:
    unordered_map<int,int>pos;
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int n=preorder.size();
        for(int i=0;i<n;i++)pos[inorder[i]]=i;
        return dfs(preorder,inorder,0,n-1,0,n-1);
    }
    TreeNode* dfs(vector<int>& preorder, vector<int>& inorder,int pl,int pr,int il,int ir) {
        if(pl>pr)
            return NULL;
        int val=preorder[pl];
        int k=pos[val];
        int len=k-il;
        auto root=new TreeNode(val);
        root->left=dfs(preorder,inorder,pl+1,pl+len,il,k-1);
        root->right=dfs(preorder,inorder,pl+len+1,pr,k+1,ir);
        return root;
    }
};