根据一棵树的中序遍历与后序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。例如,给出
中序遍历 inorder = [9,3,15,20,7] 后序遍历 postorder = [9,15,7,20,3]返回如下的二叉树:
3 / \ 9 20 / \ 15 7递归求解
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) { if(!postorder.size()) return NULL; vector<int>in_left; vector<int>in_right; vector<int>po_left; vector<int>po_right; int i,j; i=j=0; int mid=postorder[postorder.size()-1]; for(;i<inorder.size();i++) { if(inorder[i]!=mid) in_left.push_back(inorder[i]); else break; } i++; for(;i<inorder.size();i++) { in_right.push_back(inorder[i]); } for(;j<in_left.size();j++) { po_left.push_back(postorder[j]); } for(;j<postorder.size()-1;j++) { po_right.push_back(postorder[j]); } TreeNode *t=new TreeNode(mid); t->left=buildTree(in_left,po_left); t->right=buildTree(in_right,po_right); return t; } };
Leetcode 106. 从中序与后序遍历序列构造二叉树 递归
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