LeetCode #106 — MEDIUM

Construct Binary Tree from Inorder and Postorder Traversal

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Tree

Example 1

[9,3,15,20,7]
[9,15,7,20,3]

Example 2

[-1]
[-1]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #106: Construct Binary Tree from Inorder and Postorder Traversal
/**
 * 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> d = new HashMap<>();
    private int[] postorder;

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        this.postorder = postorder;
        int n = inorder.length;
        for (int i = 0; i < n; ++i) {
            d.put(inorder[i], i);
        }
        return dfs(0, 0, n);
    }

    private TreeNode dfs(int i, int j, int n) {
        if (n <= 0) {
            return null;
        }
        int v = postorder[j + n - 1];
        int k = d.get(v);
        TreeNode l = dfs(i, j, k - i);
        TreeNode r = dfs(k + 1, j + k - i, n - k + i - 1);
        return new TreeNode(v, l, r);
    }
}

// Accepted solution for LeetCode #106: Construct Binary Tree from Inorder and Postorder Traversal
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func buildTree(inorder []int, postorder []int) *TreeNode {
	d := map[int]int{}
	for i, v := range inorder {
		d[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 := d[v]
		l := dfs(i, j, k-i)
		r := dfs(k+1, j+k-i, n-k+i-1)
		return &TreeNode{v, l, r}
	}
	return dfs(0, 0, len(inorder))
}

# Accepted solution for LeetCode #106: Construct Binary Tree from Inorder and Postorder Traversal
# 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]) -> Optional[TreeNode]:
        def dfs(i: int, j: int, n: int) -> Optional[TreeNode]:
            if n <= 0:
                return None
            v = postorder[j + n - 1]
            k = d[v]
            l = dfs(i, j, k - i)
            r = dfs(k + 1, j + k - i, n - k + i - 1)
            return TreeNode(v, l, r)

        d = {v: i for i, v in enumerate(inorder)}
        return dfs(0, 0, len(inorder))

// Accepted solution for LeetCode #106: Construct Binary Tree from Inorder and Postorder Traversal
// 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::cell::RefCell;
use std::collections::HashMap;
use std::rc::Rc;
impl Solution {
    pub fn build_tree(inorder: Vec<i32>, postorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
        let n = inorder.len();
        let mut d: HashMap<i32, usize> = HashMap::new();
        for i in 0..n {
            d.insert(inorder[i], i);
        }
        fn dfs(
            postorder: &[i32],
            d: &HashMap<i32, usize>,
            i: usize,
            j: usize,
            n: usize,
        ) -> Option<Rc<RefCell<TreeNode>>> {
            if n <= 0 {
                return None;
            }
            let val = postorder[j + n - 1];
            let k = *d.get(&val).unwrap();
            let left = dfs(postorder, d, i, j, k - i);
            let right = dfs(postorder, d, k + 1, j + k - i, n - 1 - (k - i));
            Some(Rc::new(RefCell::new(TreeNode { val, left, right })))
        }
        dfs(&postorder, &d, 0, 0, n)
    }
}

// Accepted solution for LeetCode #106: Construct Binary Tree from Inorder and Postorder Traversal
/**
 * 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 = inorder.length;
    const d: Record<number, number> = {};
    for (let i = 0; i < n; i++) {
        d[inorder[i]] = i;
    }
    const dfs = (i: number, j: number, n: number): TreeNode | null => {
        if (n <= 0) {
            return null;
        }
        const v = postorder[j + n - 1];
        const k = d[v];
        const l = dfs(i, j, k - i);
        const r = dfs(k + 1, j + k - i, n - 1 - (k - i));
        return new TreeNode(v, l, r);
    };
    return dfs(0, 0, n);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.