106. Construct Binary Tree from Inorder and Postorder Traversal

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 <= 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.

Solutions

Solution 1: Hash Table + Recursion

The last node in the post-order traversal is the root node. We can find the position of the root node in the in-order traversal, and then recursively construct the left and right subtrees.

Specifically, we first use a hash table $d$ to store the position of each node in the in-order traversal. Then we design a recursive function $dfs(i, j, n)$, where $i$ and $j$ represent the starting positions of the in-order and post-order traversals, respectively, and $n$ represents the number of nodes in the subtree. The function logic is as follows:

• If $n \leq 0$, it means the subtree is empty, return a null node.
• Otherwise, take out the last node $v$ of the post-order traversal, and then find the position $k$ of $v$ in the in-order traversal using the hash table $d$. Then the number of nodes in the left subtree is $k - i$, and the number of nodes in the right subtree is $n - k + i - 1$.
• Recursively construct the left subtree $dfs(i, j, k - i)$ and the right subtree $dfs(k + 1, j + k - i, n - k + i - 1)$, connect them to the root node, and finally return the root node.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

Python3JavaC++GoTypeScriptRust

# 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))

/**
 * 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);
    }
}

/**
 * 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:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        unordered_map<int, int> d;
        int n = inorder.size();
        for (int i = 0; i < n; ++i) {
            d[inorder[i]] = i;
        }
        function<TreeNode*(int, int, int)> dfs = [&](int i, int j, int n) -> TreeNode* {
            if (n <= 0) {
                return nullptr;
            }
            int v = postorder[j + n - 1];
            int k = d[v];
            auto l = dfs(i, j, k - i);
            auto r = dfs(k + 1, j + k - i, n - k + i - 1);
            return new TreeNode(v, l, r);
        };
        return dfs(0, 0, n);
    }
};

/**
 * 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))
}

/**
 * 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);
}

// 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)
    }
}

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