【LeetCode】105. Construct Binary Tree from Preorder and Inorder Traversal

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.

可与Construct Binary Tree from Inorder and Postorder Traversal对照来看。

前序遍历Preorder的第一个节点为根，由于没有重复值，可使用根将中序Inorder划分为左右子树。

递归下去即可。

/**
 * Definition for binary tree
 * 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 Helper(preorder, , preorder.size()-, inorder, , inorder.size()-);
    }

    TreeNode* Helper(vector<int> &preorder, int begin1, int end1, vector<int> &inorder, int begin2, int end2)
    {
        if(begin1 > end1)
            return NULL;
        else if(begin1 == end1)
            return new TreeNode(preorder[begin1]);

        TreeNode* root = new TreeNode(preorder[begin1]);
        int i = begin2;
        for(; i <= end2; i ++)
        {
            if(inorder[i] == preorder[begin1])
                break;
        }
        //inorder[i] is the root
        int leftlen = i-begin2;

        //preorder[begin1] is root
        //inorder[begin2+leftlen] is root
        root->left = Helper(preorder, begin1+, begin1+leftlen, inorder, begin2, begin2+leftlen-);
        root->right = Helper(preorder, begin1+leftlen+, end1, inorder, begin2+leftlen+, end2);
        return root;
    }
};