输入某二叉树的前序遍历和中序遍历的结果，请构建该二叉树并返回其根节点。
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。
示例 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] Output: [3,9,20,null,null,15,7]
示例 2:
Input: preorder = [-1], inorder = [-1] Output: [-1]
限制：
0 <= 节点个数 <= 5000
注意：本题与主站 105 题重复：https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
前序序列的第一个结点 preorder[0] 为根节点，我们在中序序列中找到根节点的位置 i，可以将中序序列划分为左子树 inorder[:i] 和右子树 inorder[i + 1:]。
通过左右子树的区间，可以计算出左、右子树节点的个数，假设为 m、n。然后在前序节点中，从根节点往后的 m 个节点为左子树，再往后的 n 个节点为右子树。
递归求解即可。
前序遍历：先遍历根节点，再遍历左子树，最后遍历右子树；
中序遍历：先遍历左子树，再遍历根节点，最后遍历右子树。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode:
if not preorder:
return None
v = preorder[0]
root = TreeNode(val=v)
i = inorder.index(v)
root.left = self.buildTree(preorder[1:1 + i], inorder[:i])
root.right = self.buildTree(preorder[1 + i:], inorder[i + 1:])
return root/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private Map<Integer, Integer> indexes = new HashMap<>();
public TreeNode buildTree(int[] preorder, int[] inorder) {
for (int i = 0; i < inorder.length; ++i) {
indexes.put(inorder[i], i);
}
return dfs(preorder, inorder, 0, 0, preorder.length);
}
private TreeNode dfs(int[] preorder, int[] inorder, int i, int j, int n) {
if (n <= 0) {
return null;
}
int v = preorder[i];
int k = indexes.get(v);
TreeNode root = new TreeNode(v);
root.left = dfs(preorder, inorder, i + 1, j, k - j);
root.right = dfs(preorder, inorder, i + 1 + k - j, k + 1, n - k + j - 1);
return root;
}
}/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {number[]} preorder
* @param {number[]} inorder
* @return {TreeNode}
*/
var buildTree = function (preorder, inorder) {
if (preorder.length == 0) return null;
const v = preorder[0];
const root = new TreeNode(v);
const i = inorder.indexOf(v);
root.left = buildTree(preorder.slice(1, 1 + i), inorder.slice(0, i));
root.right = buildTree(preorder.slice(1 + i), inorder.slice(1 + i));
return root;
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func buildTree(preorder []int, inorder []int) *TreeNode {
indexes := make(map[int]int)
for i, v := range inorder {
indexes[v] = i
}
var dfs func(i, j, n int) *TreeNode
dfs = func(i, j, n int) *TreeNode {
if n <= 0 {
return nil
}
v := preorder[i]
k := indexes[v]
root := &TreeNode{Val: v}
root.Left = dfs(i+1, j, k-j)
root.Right = dfs(i+1+k-j, k+1, n-k+j-1)
return root
}
return dfs(0, 0, len(inorder))
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
unordered_map<int, int> indexes;
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
for (int i = 0; i < inorder.size(); ++i) indexes[inorder[i]] = i;
return dfs(preorder, inorder, 0, 0, inorder.size());
}
TreeNode* dfs(vector<int>& preorder, vector<int>& inorder, int i, int j, int n) {
if (n <= 0) return nullptr;
int v = preorder[i];
int k = indexes[v];
TreeNode* root = new TreeNode(v);
root->left = dfs(preorder, inorder, i + 1, j, k - j);
root->right = dfs(preorder, inorder, i + 1 + k - j, k + 1, n - k + j - 1);
return root;
}
};/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
if (preorder.length == 0) return null;
let val: number = preorder[0];
let node: TreeNode = new TreeNode(val);
let index: number = inorder.indexOf(val);
node.left = buildTree(
preorder.slice(1, index + 1),
inorder.slice(0, index),
);
node.right = buildTree(preorder.slice(index + 1), inorder.slice(index + 1));
return node;
}/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function buildTree(preorder: number[], inorder: number[]): TreeNode | null {
if (inorder.length === 0) {
return null;
}
const val = preorder[0];
const i = inorder.indexOf(val);
return new TreeNode(
val,
buildTree(preorder.slice(1, i + 1), inorder.slice(0, i)),
buildTree(preorder.slice(i + 1), inorder.slice(i + 1)),
);
}// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
pub fn build_tree(preorder: Vec<i32>, inorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
if inorder.is_empty() {
return None;
}
let val = preorder[0];
let i = inorder.iter().position(|num| *num == val).unwrap();
Some(Rc::new(RefCell::new(TreeNode {
val,
left: Self::build_tree(preorder[1..i + 1].to_vec(), inorder[..i].to_vec()),
right: Self::build_tree(preorder[i + 1..].to_vec(), inorder[i + 1..].to_vec()),
})))
}
}