LeetCode

根据二叉树的前序遍历和中序遍历构造二叉树。

思路：前序遍历的第一个节点就是根节点，扫描中序遍历找出根结点，根结点的左边、右边分别为左子树、右子树中序遍历。再计算左子数长度leftLength，前序遍历根结点后的leftLength长度为左子树的前序遍历，剩下的为右子树的前序遍历，代码如下：

/**
 * 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)
    {
        if (preorder.size() == 0 || inorder.size() == 0)
            return NULL;
        
        return build(&preorder[0], &preorder[preorder.size()-1], &inorder[0], &inorder[inorder.size()-1]);
    }
    TreeNode *build(int *preStart, int *preEnd, int *inStart, int *inEnd)
    {
        //前序遍历的第一个值为根结点的
        int rootValue = *preStart;
        TreeNode *root = new TreeNode(rootValue);
        
        if (preStart == preEnd)
        {
            if (inStart == inEnd && *preStart == *inStart)
                return root;
            else
                cout << "Invalid input." << endl;
        }
        
        //在中序遍历中找到根结点的值
        int *rootInorder = inStart;
        while (rootInorder <= inEnd && *rootInorder != rootValue)
            ++rootInorder;
        if (rootInorder == inEnd && *rootInorder != rootValue)
            cout << "Invalid input." << endl;
            
        int leftLength = rootInorder - inStart;
        int *leftPreEnd = preStart + leftLength;
        if (leftLength > 0)
        {
            //构建左子树
            root->left = build(preStart+1, leftPreEnd, inStart, rootInorder-1);
        }
        if (leftLength < preEnd - preStart)
        {
            //构建右子
            root->right = build(leftPreEnd+1, preEnd, rootInorder+1, inEnd);
        }
        return root;
    }
};