Given two integer arrays preorder
and inorder
where preorder
is the preorder traversal of a binary tree and inorder
is the inorder traversal of the same tree, construct and return the binary tree.
Example 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7] Output: [3,9,20,null,null,15,7]
Example 2:
Input: preorder = [-1], inorder = [-1] Output: [-1]
Constraints:
1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder
andinorder
consist of unique values.- Each value of
inorder
also appears inpreorder
. preorder
is guaranteed to be the preorder traversal of the tree.inorder
is guaranteed to be the inorder traversal of the tree.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
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() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
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.
* 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> 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.
* 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))
}