Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
Solution:
/** * 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 *built(vector<int> &inorder, vector<int> &postorder, int in_start, int in_end, int post_start, int post_end) { if(in_start > in_end || post_start > post_end) return NULL; TreeNode *curRoot = new TreeNode(postorder[post_end]); int rootIndex = -1; for(int i = in_end;i >= in_start;i--) { if(inorder[i] == postorder[post_end]) { rootIndex = i; break; } } if(rootIndex == -1) return NULL; int leftNum = rootIndex - in_start; curRoot -> left = built(inorder, postorder, in_start, rootIndex - 1, post_start, post_start + leftNum - 1); curRoot -> right = built(inorder, postorder, rootIndex + 1, in_end, post_start + leftNum, post_end - 1); return curRoot; } TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) { return built(inorder, postorder, 0, inorder.size() - 1, 0, postorder.size() - 1); } };