[Leetcode] 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.

没什么好说的，了解原理即可，使用递归。

/**
 * 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, int a, int b, int n) {
        TreeNode *root = NULL;
        if (n < 1) {
            return root;
        }
        root = new TreeNode(preorder[a]);
        bool flag = false;
        int i;
        for (i = 0; i < n; ++i) {
            if (inorder[b+i] == preorder[a]) {
                break;
            }
        }
        root->left = buildTree(preorder, inorder, a+1, b, i);
        root->right = buildTree(preorder, inorder, a+i+1, b+i+1, n-1-i);
        return root;
    }

    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        int n = preorder.size();
        return buildTree(preorder, inorder, 0, 0, n);
        
    }
};