# Question

Formatted question description: https://leetcode.ca/all/106.html

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 <= 3000
• postorder.length == inorder.length
• -3000 <= inorder[i], postorder[i] <= 3000
• inorder 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.

# Algorithm

The last one in the post-order sequence must be the root, so the root node of the original binary tree can be known. A very key condition is given in the title that there are no identical elements in the tree.

With this condition, we can also locate in the in-order traversal to find the position of the root node, and split the in-order traversal into the left and right parts according to the position of the root node, and call the original function on it recursively

# Code

• import java.util.Arrays;

public class Construct_Binary_Tree_from_Inorder_and_Postorder_Traversal {

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/

public class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {

return buildTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
}
TreeNode buildTree(int[] inorder, int iLeft, int iRight, int[] postorder, int pLeft, int pRight) {
if (iLeft > iRight || pLeft > pRight) return null;
TreeNode cur = new TreeNode(postorder[pRight]);
int i = 0;
for (i = iLeft; i < inorder.length; ++i) {
if (inorder[i] == cur.val) break;
}
cur.left = buildTree(inorder, iLeft, i - 1, postorder, pLeft, pLeft + (i - iLeft - 1));
cur.right = buildTree(inorder, i + 1, iRight, postorder, pLeft + (i - iLeft), pRight - 1); // (i - iLeft) is (i - iLeft - 1 + 1)
return cur;
}
}
}

############

/**
* 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;
}
}

• // OJ: https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
// Time: O(N^2)
// Space: O(H)
class Solution {
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
function<TreeNode*(int, int, int, int)> build = [&](int is, int ie, int ps, int pe) -> TreeNode* {
if (is == ie) return NULL;
auto n = new TreeNode(postorder[pe - 1]);
int mid = find(begin(inorder) + is, begin(inorder) + ie, n->val) - begin(inorder);
n->left = build(is, mid, ps, ps + mid - is);
n->right = build(mid + 1, ie, ps + mid - is, pe - 1);
return n;
};
return build(0, inorder.size(), 0, postorder.size());
}
};

• # 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

#############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

'''
search for index, also use .index(val)
>>> a = [1,2,3,4,5]
>>> a.index(3)
2
'''
class Solution(object):
def buildTree(self, inorder, postorder):
"""
:type inorder: List[int]
:type postorder: List[int]
:rtype: TreeNode
"""
if inorder and postorder:
postorder.reverse()
self.index = 0
d = {}
for i in range(0, len(inorder)):
d[inorder[i]] = i
return self.dfs(inorder, postorder, 0, len(postorder) - 1, d)

def dfs(self, inorder, postorder, start, end, d):
if start <= end:
root = TreeNode(postorder[self.index])
mid = d[postorder[self.index]]
self.index += 1
root.right = self.dfs(inorder, postorder, mid + 1, end, d)
root.left = self.dfs(inorder, postorder, start, mid - 1, d)
return root


• /**
* 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))
}

• /**
* 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)),
);
}


• // 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)
}
}