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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: 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.

  • /**
 * 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;
    }
}

  • /**
 * 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;
    }
};

  • # 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)
    }
}

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