106. Construct Binary Tree from Inorder and Postorder Traversal (Tree; DFS)
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
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) { // Start typing your C/C++ solution below // DO NOT write int main() function root = NULL; if(inorder.empty()) return root; root = new TreeNode(0); buildSubTree(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1,root); return root; } void buildSubTree(vector<int> &inorder, vector<int>&postorder, int inStartPos, int inEndPos, int postStartPos, int postEndPos,TreeNode * currentNode) { currentNode->val = postorder[postEndPos]; //后序遍历的最后一个节点是根节点 //find root position in inorder vector int inRootPos; for(int i = inStartPos; i <= inEndPos; i++) { if(inorder[i] == postorder[postEndPos]) { inRootPos = i; break; } } //right tree: 是中序遍历根节点之后的部分,对应后序遍历根节点前相同长度的部分 int newPostPos = postEndPos - max(inEndPos - inRootPos, 0); if(inRootPos<inEndPos) { currentNode->right = new TreeNode(0); buildSubTree(inorder,postorder,inRootPos+1,inEndPos,newPostPos,postEndPos-1,currentNode->right); } //leftTree: 是中序遍历根节点之前的部分,对应后序遍历从头开始相同长度的部分 if(inRootPos>inStartPos) { currentNode->left = new TreeNode(0); buildSubTree(inorder,postorder,inStartPos,inRootPos-1,postStartPos,newPostPos-1,currentNode->left); } } private: TreeNode* root; };