题目描述
输入某二叉树的前序遍历和中序遍历的结果,请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6},则重建二叉树并返回。
解题思路
常规题型,根据二叉树的前序,中序结果可以完全确定一棵树,根据前序遍历的第一个数字创建根节点,在中序遍历中找到根节点的位置,由该位置可以划分右子树和左子树的序列,接着递归调用函数构建左右子树
C++版
class Solution {
public:
TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
if(pre.size() == 0)
return nullptr;
vector<int>preLeft, preRight, inLeft, inRight;
TreeNode *head = new TreeNode(pre[0]);
int root = 0;
for(int i = 0; i < pre.size(); ++ i)
if(pre[0] == vin[i]){
root = i;
break;
}
for(int i = 0; i < root; ++ i){
inLeft.push_back(vin[i]);
preLeft.push_back(pre[i+1]);
}
for(int i = root + 1; i < pre.size(); ++ i){
inRight.push_back(vin[i]);
preRight.push_back(pre[i]);
}
head->left = reConstructBinaryTree(preLeft, inLeft);
head->right = reConstructBinaryTree(preRight, inRight);
return head;
}
};
Python版
class Solution:
# 返回构造的TreeNode根节点
def reConstructBinaryTree(self, pre, tin):
# write code here
if len(pre) == 0:
return None
elif len(pre) == 1:
return TreeNode(pre[0])
root = TreeNode(pre[0])
pos = tin.index(pre[0])
root.left = self.reConstructBinaryTree(pre[1:pos+1], tin[:pos])
root.right = self.reConstructBinaryTree(pre[pos+1:], tin[pos+1:])
return root