LeetCode: Construct Binary Tree from Inorder and Postorder Traversal

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

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

地址：https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

算法：根据中序和后序构造出二叉树。细心一点应该不会错。代码：

/**
* 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 *buildTree(vector<int> &inorder, vector<int> &postorder) {
if(inorder.empty()) return NULL;
return subBuildTree(inorder,postorder,0,inorder.size()-1,0,postorder.size()-1);
}
TreeNode *subBuildTree(vector<int> &inorder, vector<int> &postorder, int inbegin, int inend, int postbegin, int postend){
TreeNode *root = new TreeNode(postorder[postend]);
if(inbegin == inend && postbegin == postend){
return root;
}
int key = postorder[postend];
int i = inbegin;
while(i <= inend && inorder[i] != key)  ++i;
if(i > inbegin){
root->left = subBuildTree(inorder,postorder,inbegin,i-1,postbegin,postbegin+i-1-inbegin);
}
if(inend > i){
root->right = subBuildTree(inorder,postorder,i+1,inend,postbegin+i-inbegin,postend-1);
}
return root;
}
};