[LeetCode] Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
递归构造
1 /** 2 * Definition for binary tree 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 *createTree(vector<int> &inorder, int inBeg, int inEnd, vector<int> &postorder, int postBeg, int postEnd) 13 { 14 if (inBeg > inEnd) 15 return NULL; 16 17 int root = postorder[postEnd]; 18 19 int index; 20 21 for(int i = inBeg; i <= inEnd; i++) 22 if (inorder[i] == root) 23 { 24 index = i; 25 break; 26 } 27 28 int len = index - inBeg; 29 TreeNode *left = createTree(inorder, inBeg, index - 1, postorder, postBeg, postBeg + len - 1); 30 TreeNode *right = createTree(inorder, index + 1, inEnd, postorder, postBeg + len, postEnd - 1); 31 32 TreeNode *node = new TreeNode(root); 33 node->left = left; 34 node->right = right; 35 36 return node; 37 } 38 39 TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { 40 // Start typing your C/C++ solution below 41 // DO NOT write int main() function 42 if (inorder.size() == 0) 43 return NULL; 44 45 TreeNode *head = createTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1); 46 47 return head; 48 } 49 };
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