[LeetCode] 020: Construct Binary Tree from Inorder and Postorder Traversal

[Problem]
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:
// search x in array[begin:end]
int search(vector<int> &array, int begin, int end, int x){
for(int i = begin; i <= end; ++i){
if(array[i] == x)return i;
}
return -1;
}
// build tree with inorder[begin1:end1] and postorder[begin2:end2]
TreeNode *buildTree(vector<int> &inorder, int begin1, int end1, vector<int> &postorder, int begin2, int end2){
// invalid
if(begin1 > end1 || begin2 > end2 || end1 - begin1 != end2 - begin2){
return NULL;
}
if(begin1 == end1){
// invalid
if(inorder[begin1] != postorder[begin2]){
return NULL;
}
return new TreeNode(inorder[begin1]);
}

// search the root in the inorder
int mid = search(inorder, begin1, end1, postorder[end2]);
if(mid == -1) return NULL;

// build the left brunch
TreeNode *left = buildTree(inorder, begin1, mid-1, postorder, begin2, begin2 + (mid - begin1) - 1);

// build the right brunch
TreeNode *right = buildTree(inorder, mid+1, end1, postorder, begin2 + (mid - begin1), end2 - 1);

// set the left brunch and the right brunch
TreeNode *root = new TreeNode(postorder[end2]);
root->left = left;
root->right = right;
return root;
}
// build tree with inorder and postorder
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return buildTree(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);
}
};

说明：版权所有，转载请注明出处。Coder007的博客