标签:solution traversal public dtree ram 中序 size empty 前序遍历
根据中序遍历和后序遍历树构造二叉树
样例:
给出树的中序遍历: [1,2,3] 和后序遍历: [1,3,2]
返回如下的树:
2
/ \
1 3
借鉴上一篇《前序遍历和中序遍历树构造二叉树》,我们知道中序遍历为左->中->右,后序遍历为左->右->中。于是后序遍历的最后一个值即为根节点的值,根据这个值我们在中序遍历中找到根节点左子树和右子树的值,递归构造左子树和右子树即可。
/** * 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 inorder : A list of integers that inorder traversal of a tree *@param postorder : A list of integers that postorder traversal of a tree *@return : Root of a tree */ public: TreeNode *construct(vector<int> &inorder, vector<int> &postorder, int is, int ie, int ps, int pe) { TreeNode * root = new TreeNode(postorder[pe]); if (ps == pe) return root; int i; for (i = 0; i < ie; i++) { if (inorder[i] == root->val) break; } if (i-1 >= is) { root->left = construct(inorder, postorder, is, i-1, ps, ps+i-1-is); } if (i+1 <= ie) { root->right = construct(inorder, postorder, i+1, ie, ps+i-is, pe-1); } return root; } TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { if (inorder.empty() || postorder.empty() || inorder.size() != postorder.size()) return NULL; return construct(inorder, postorder, 0, inorder.size()-1, 0, postorder.size()-1); } };
标签:solution traversal public dtree ram 中序 size empty 前序遍历
原文地址:http://www.cnblogs.com/lingd3/p/7073671.html