描述
根据中序遍历和后序遍历树构造二叉树
你可以假设树中不存在相同数值的节点
您在真实的面试中是否遇到过这个题? 是
样例
给出树的中序遍历: [1,2,3] 和后序遍历: [1,3,2]
返回如下的树:
2
/ \
1 3
题目链接
程序
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
public:
/**
* @param inorder: A list of integers that inorder traversal of a tree
* @param postorder: A list of integers that postorder traversal of a tree
* @return: Root of a tree
*/
TreeNode * buildTree(vector<int> &inorder, vector<int> &postorder) {
// write your code here
TreeNode * root = NULL;
vector<int> l_postorder, r_postorder, l_inorder, r_inorder;
int i, root_index = 0;
if(postorder.empty() != 1 || inorder.empty() != 1){
// 在后序队列中找根节点
root = new TreeNode(postorder[postorder.size() - 1]);
// 在中序队列中找出根节点位置
for(i = 0; i < inorder.size(); i++){
if(postorder[postorder.size() - 1] == inorder[i])
break;
root_index++;
}
// 左右子树的后序、中序队列
for(i = 0; i < root_index; i++){
l_inorder.push_back(inorder[i]);
l_postorder.push_back(postorder[i]);
}
for(i = root_index + 1; i < inorder.size(); i++){
r_inorder.push_back(inorder[i]);
r_postorder.push_back(postorder[i-1]);
}
root->left = buildTree(l_inorder, l_postorder);
root->right = buildTree(r_inorder, r_postorder);
}
return root;
}
};