Description#
Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.
Example 1:
Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
Output: [3,9,20,null,null,15,7]
Example 2:
Input: inorder = [-1], postorder = [-1]
Output: [-1]
Constraints:
1 <= inorder.length <= 3000postorder.length == inorder.length-3000 <= inorder[i], postorder[i] <= 3000inorder and postorder consist of unique values.- Each value of
postorder also appears in inorder. inorder is guaranteed to be the inorder traversal of the tree.postorder is guaranteed to be the postorder traversal of the tree.
Solutions#
Solution 1: Recursion#
The approach is the same as in 105. Construct Binary Tree from Preorder and Inorder Traversal.
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.
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| # 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, inorder: List[int], postorder: List[int]) -> TreeNode:
if not postorder:
return None
v = postorder[-1]
root = TreeNode(val=v)
i = inorder.index(v)
root.left = self.buildTree(inorder[:i], postorder[:i])
root.right = self.buildTree(inorder[i + 1 :], postorder[i:-1])
return root
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| /**
* 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[] inorder, int[] postorder) {
for (int i = 0; i < inorder.length; ++i) {
indexes.put(inorder[i], i);
}
return dfs(inorder, postorder, 0, 0, inorder.length);
}
private TreeNode dfs(int[] inorder, int[] postorder, int i, int j, int n) {
if (n <= 0) {
return null;
}
int v = postorder[j + n - 1];
int k = indexes.get(v);
TreeNode root = new TreeNode(v);
root.left = dfs(inorder, postorder, i, j, k - i);
root.right = dfs(inorder, postorder, k + 1, j + k - i, n - k + i - 1);
return root;
}
}
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| /**
* 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>& inorder, vector<int>& postorder) {
for (int i = 0; i < inorder.size(); ++i) indexes[inorder[i]] = i;
return dfs(inorder, postorder, 0, 0, inorder.size());
}
TreeNode* dfs(vector<int>& inorder, vector<int>& postorder, int i, int j, int n) {
if (n <= 0) return nullptr;
int v = postorder[j + n - 1];
int k = indexes[v];
TreeNode* root = new TreeNode(v);
root->left = dfs(inorder, postorder, i, j, k - i);
root->right = dfs(inorder, postorder, k + 1, j + k - i, n - k + i - 1);
return root;
}
};
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| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func buildTree(inorder []int, postorder []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 := postorder[j+n-1]
k := indexes[v]
root := &TreeNode{Val: v}
root.Left = dfs(i, j, k-i)
root.Right = dfs(k+1, j+k-i, n-k+i-1)
return root
}
return dfs(0, 0, len(inorder))
}
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| /**
* 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(inorder: number[], postorder: number[]): TreeNode | null {
const n = postorder.length;
if (n == 0) {
return null;
}
const val = postorder[n - 1];
const index = inorder.indexOf(val);
return new TreeNode(
val,
buildTree(inorder.slice(0, index), postorder.slice(0, index)),
buildTree(inorder.slice(index + 1), postorder.slice(index, n - 1)),
);
}
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| // 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 {
fn reset(
inorder: &Vec<i32>,
i_left: usize,
i_right: usize,
postorder: &Vec<i32>,
p_left: usize,
p_right: usize
) -> Option<Rc<RefCell<TreeNode>>> {
if i_left == i_right {
return None;
}
let val = postorder[p_right - 1];
let index = inorder
.iter()
.position(|&v| v == val)
.unwrap();
Some(
Rc::new(
RefCell::new(TreeNode {
val,
left: Self::reset(
inorder,
i_left,
index,
postorder,
p_left,
p_left + index - i_left
),
right: Self::reset(
inorder,
index + 1,
i_right,
postorder,
p_left + index - i_left,
p_right - 1
),
})
)
)
}
pub fn build_tree(inorder: Vec<i32>, postorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
let n = inorder.len();
Self::reset(&inorder, 0, n, &postorder, 0, n)
}
}
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