从前序与中序遍历序列构造二叉树 🌟🌟🌟🌟🌟中等
课后作业
问题描述
原文链接：105. 从前序与中序遍历序列构造二叉树
给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。
示例 1:
输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]
示例 2:
输入: preorder = [-1], inorder = [-1]
输出: [-1]
提示:
1 <= preorder.length <= 3000inorder.length == preorder.length-3000 <= preorder[i], inorder[i] <= 3000preorder和inorder均无重复元素inorder均出现在preorderpreorder保证为二叉树的前序遍历序列inorder保证为二叉树的中序遍历序列
代码实现
Java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
Map<Integer, Integer> map = new HashMap();
public TreeNode buildTree(int[] preorder, int[] inorder) {
for(int i = 0; i < inorder.length; i++){
map.put(inorder[i], i);
}
return f(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
}
TreeNode f(int[] pre, int l1, int r1, int[] in, int l2, int r2){
if(l1 > r1 || l2 > r2){
return null;
}
TreeNode root = new TreeNode(pre[l1]);
int i = map.get(pre[l1]);
root.left = f(pre, l1+1, l1+(i-l2), in, l2, i-1);
root.right = f(pre, l1+(i-l2)+1, r1, in, i + 1, r2);
return root;
}
}
Python
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution(object):
def buildTree(self, preorder, inorder):
"""
:type preorder: List[int]
:type inorder: List[int]
:rtype: TreeNode
"""
map = {}
for i in range(len(inorder)):
map[inorder[i]] = i
return self.f(preorder, 0, len(preorder)-1, inorder, 0, len(inorder)-1, map)
def f(self, pre, l1, r1, ino, l2, r2, map):
if l1 > r1 or l2 > r2:
return None
root = TreeNode(pre[l1])
index = map[pre[l1]]
root.left = self.f(pre, l1+1, l1+(index-l2), ino, l2, index-1, map)
root.right = self.f(pre, l1+(index-l2)+1, r1, ino, index+1, r2, map)
return root
C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
unordered_map<int, int> map;
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
for (int i = 0; i < inorder.size(); i++) {
map[inorder[i]] = i;
}
return f(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
}
TreeNode* f(vector<int>& pre, int l1, int r1, vector<int>& in, int l2, int r2) {
if (l1 > r1 || l2 > r2) {
return nullptr;
}
TreeNode* root = new TreeNode(pre[l1]);
int index = map[pre[l1]];
root->left = f(pre, l1+1, l1+(index-l2), in, l2, index-1);
root->right = f(pre, l1+(index-l2)+1, r1, in, index+1, r2);
return root;
}
};
Go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func buildTree(preorder []int, inorder []int) *TreeNode {
hashMap := make(map[int]int)
for i := 0; i < len(inorder); i++ {
hashMap[inorder[i]] = i
}
return f(preorder, 0, len(preorder)-1, inorder, 0, len(inorder)-1, hashMap)
}
func f(pre []int, l1 int, r1 int, ino []int, l2 int, r2 int, hashMap map[int]int) *TreeNode {
if l1 > r1 || l2 > r2 {
return nil
}
root := &TreeNode{Val: pre[l1]}
index := hashMap[pre[l1]]
root.Left = f(pre, l1+1, l1+(index-l2), ino, l2, index-1, hashMap)
root.Right = f(pre, l1+(index-l2)+1, r1, ino, index+1, r2, hashMap)
return root
}