Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* build(vector<int> &inorder,int left1,int right1,vector<int> &postorder,int left2,int right2) { if(right1-left1 != right2-left2)return NULL; if(right1>=inorder.size() || right2>=postorder.size())return NULL; if(left1==right1 && left2==right2) { TreeNode* root = new TreeNode(inorder[left1]); return root; }else if(left1<right1 && left2<right2) { TreeNode* root = new TreeNode(postorder[right2]); int i; for(i = right1; i >= left1; i--) if(inorder[i] == postorder[right2])break; if(i<left1)return NULL; root->left = build(inorder,left1,i-1,postorder,left2,right2+i-right1-1); root->right = build(inorder,i+1,right1,postorder,right2+i-right1,right2-1); return root; }else return NULL; } TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { // Note: The Solution object is instantiated only once. return build(inorder,0,inorder.size()-1,postorder,0,postorder.size()-1); } };