[leetcode] 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.

思路：

由先序遍历和中序遍历确定二叉树的结构。还是那个问题，index-l+i+1是我试出来的，具体为什么需要好好想想。

题解：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void build(vector<int> &preorder, vector<int> &inorder, int l, int r, int index, TreeNode *&root) {
        if(l>r)
            return;
        root = new TreeNode(preorder[index]);
        int i;
        for(i=l;i<=r;i++)
            if(inorder[i]==preorder[index])
                break;
        build(preorder, inorder, l, i-1, index+1, root->left);
        build(preorder, inorder, i+1, r, index-l+i+1, root->right);
    }
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        if(preorder.size()==0)
            return NULL;
        TreeNode *root;
        build(preorder, inorder, 0, inorder.size()-1, 0, root);
        return root;
    }
};