前序遍历和中序遍历树构造二叉树

根据前序遍历和中序遍历树构造二叉树

样例:

给出中序遍历：[1,2,3]和前序遍历：[2,1,3]. 返回如下的树:

  2

 /  \

1  3

我们知道前序遍历是中->左->右，中序遍历是左->中->右。因此根据前序遍历的第一个数，即为根节点，我们可以在中序遍历中找到根节点的左子树和右子树，同样递归在左子树中找到左子树的根节点和其左右子树，对右子树也一样。这样理清思路后代码就不难写出来了。

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */ class Solution {    /**     *@param preorder : A list of integers that preorder traversal of a tree     *@param inorder : A list of integers that inorder traversal of a tree     *@return : Root of a tree     */public:    TreeNode *construct(vector<int> &preorder, vector<int> &inorder, int ps,         int pe, int is, int ie) {            TreeNode * root = new TreeNode(preorder[ps]);            if (ps == pe) return root;            int i;            for (i = 0; i < ie; i++) {                   if (inorder[i] == root->val) break;   //找到根节点位置            }            //递归构造左右子树            if (is <= i-1)                 root->left = construct(preorder, inorder,                     ps+1, ps+(i-is), is, i-1);            if (i+1 <= ie)                 root->right = construct(preorder, inorder,                    ps+(i-is)+1, pe, i+1, ie);            return root;        }    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {        if (preorder.empty() || inorder.empty() ||             preorder.size() != inorder.size()) return NULL;        return construct(preorder, inorder, 0, preorder.size()-1, 0, inorder.size()-1);    }};