105. Construct Binary Tree from Preorder and Inorder Traversal (递归)

---

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

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

class Solution {
private:
    unordered_map<int,int> index_map;
public:
    TreeNode* build(int pre_left,int pre_right, vector<int>& preorder, int in_left, int in_right,vector<int>& inorder) {
        if(pre_left > pre_right || in_left > in_right) return nullptr; 
        int inorder_index = index_map[preorder[pre_left]];
        int left_length = inorder_index  - in_left;
        TreeNode* root = new TreeNode(preorder[pre_left]);
        root->left = build(pre_left+1,pre_left+left_length,preorder,in_left,inorder_index-1,inorder);
        root->right= build(pre_left+left_length+1,pre_right,preorder,inorder_index+1,in_right,inorder);
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        for(int i = 0; i < inorder.size(); ++i) {
            index_map[inorder[i]] = i;
        }
        return build(0,preorder.size()-1,preorder,0,inorder.size()-1,inorder);
    }
};

 

 

 

 

 1 class Solution {
 2     public TreeNode buildTree(int[] preorder, int[] inorder) {
 3         return help(preorder,0,preorder.length-1,inorder,0,inorder.length-1);
 4     }
 5     private TreeNode help(int[] preorder,int pre_start,int pre_end,int[] inorder,int in_start,int in_end){
 6         if(in_start>in_end || pre_start>pre_end) return null;
 7         
 8         
 9         TreeNode root = new TreeNode(preorder[pre_start]);
10         int j = in_start;
11         while(preorder[pre_start]!=inorder[j]&& j <=pre_end)
12             j++;
13         //注意下标！！！！！！！！
14         root.left = help(preorder,pre_start+1,pre_start+(j-in_start),inorder,in_start,j-1);
15         root.right = help(preorder,pre_start+(j-in_start)+1,pre_end,inorder,j+1,in_end);
16         return root;
17     }
18 }