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。
思路:使用递归。前序遍历的第一个节点即为根节点,在中序遍历中,以前序遍历的第一个节点为分界线,左边的为左子树的节点,右边的为右子树的节点。使用递归,即可求解。最终返回根节点即可。
C++实现:
/**
* 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* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
int size = vin.size();
if (size == 0)
return NULL;
int val = pre[0], g = 0;
TreeNode *T = new TreeNode(val);
vector <int> pre_left, pre_right, vin_left, vin_right;
for(int i = 0; i < size; i ++)
{
if(vin[i] == val)
{
g = i;
break;
}
}
for(int i = 0; i < g; i ++)
{
pre_left.push_back(pre[i + 1]);
vin_left.push_back(vin[i]);
}
for(int i = g + 1; i < size; i ++)
{
pre_right.push_back(pre[i]);
vin_right.push_back(vin[i]);
}
(*T).left = reConstructBinaryTree(pre_left, vin_left);
(*T).right = reConstructBinaryTree(pre_right, vin_right);
return T;
}
};
python实现:
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回构造的TreeNode根节点
def reConstructBinaryTree(self, pre, tin):
# write code here
t = len(tin)
if t == 0:
return None
root = TreeNode(pre[0])
root.left = self.reConstructBinaryTree(pre[1:tin.index(pre[0]) + 1], tin[0:tin.index(pre[0])])
root.right = self.reConstructBinaryTree(pre[tin.index(pre[0]) + 1 : ], tin[tin.index(pre[0]) + 1:])
return root