标签:const cto span 构造二叉树 str 返回 tor node uil
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); } };
标签:const cto span 构造二叉树 str 返回 tor node uil
原文地址:http://www.cnblogs.com/lingd3/p/7073669.html