LeetCode OJ：Construct Binary Tree from Preorder and Inorder Traversal（从前序以及中序遍历结果中构造二叉树）

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

Note:
You may assume that duplicates do not exist in the tree.
从前序以及中序的结果中构造二叉树，这里保证了不会有两个相同的数字，用递归构造就比较方便了：

class Solution {
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if(!preorder.size()) return NULL;
        return createTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);//边界条件应该想清楚
    }

    TreeNode* createTree(vector<int>& preOrder, int preBegin, int preEnd, vector<int>& inOrder, int inBegin, int inEnd)
    {
        if(preBegin > preEnd) return NULL;
        int rootVal = preOrder[preBegin];
        int mid;
        for(int i = inBegin; i <= inEnd; ++i)
            if(inOrder[i] == rootVal){
                mid = i;
                break;
            }
        int len = mid - inBegin;    //左边区间的长度为mid - inBegin;
        TreeNode * left = createTree(preOrder, preBegin + 1, preBegin + len, inOrder, inBegin, mid - 1);
        TreeNode * right = createTree(preOrder, preBegin + len + 1, preEnd, inOrder, mid + 1, inEnd);
        TreeNode * root = new TreeNode(rootVal);
        root->left = left;
        root->right = right;
        return root;
    }
};

 java版本的代码如下所示，方法上与上面的没什么区别：

public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return createTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
    }
    
    public TreeNode createTree(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd){
        if(preBegin > preEnd)
            return null;
        int rootVal = preorder[preBegin];
        int i = inBegin;
        for(; i <= inEnd; ++i){
            if(inorder[i] == rootVal)
                break;
        }
        int leftLen = i - inBegin;
        TreeNode root = new TreeNode(rootVal);
        root.left = createTree(preorder, preBegin + 1, preBegin + leftLen ,inorder, inBegin, i - 1); //这里的边界条件应该注意
        root.right = createTree(preorder, preBegin + 1 + leftLen, preEnd, inorder, i + 1, inEnd);
        return root;
    }
}