面试题07-I. 根据中序和后序重建二叉树
作者:互联网
题目:
解答:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 TreeNode *buildTree(vector<int> &iorder, int isi, int iei, vector<int> &porder, int psi, int pei) 13 { 14 if(iei - isi < 0 || iei - isi != pei - psi) 15 { 16 return NULL; 17 } 18 19 //the porder[pei] is the root of this tree 20 21 TreeNode *root = new TreeNode(porder[pei]); 22 23 //find the root in the iorder to seperate it into left sub tree and right sub tree 24 //root index in inorder array 25 26 int riii = -1; 27 for(int i = isi; i <= iei; ++i) 28 { 29 if(iorder[i] == root->val) 30 { 31 riii = i; 32 break; 33 } 34 } 35 if(riii == -1) 36 { 37 return root;//error 38 } 39 40 int lnodes = riii - isi; 41 42 //for the left sub tree 43 //the isi to riii - 1 in inorder array will be it's inorder traversal 44 //and the psi to psi + lnodes - 1 in postorder array will be it's post order traversal 45 46 root->left = buildTree(iorder, isi, riii - 1, porder, psi, psi + lnodes - 1); 47 48 //for the right sub tree is similary to the left 49 50 root->right = buildTree(iorder, riii + 1, iei, porder, psi + lnodes, pei - 1); 51 52 return root; 53 } 54 TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) 55 { 56 return buildTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() -1); 57 } 58 };
标签:面试题,psi,TreeNode,07,int,riii,isi,二叉树,root 来源: https://www.cnblogs.com/ocpc/p/12856591.html