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105. Construct Binary Tree from Preorder and Inorder Traversal

中文文档

Description

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

 

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]
Output: [-1]

 

Constraints:

  • 1 <= preorder.length <= 3000
  • inorder.length == preorder.length
  • -3000 <= preorder[i], inorder[i] <= 3000
  • preorder and inorder consist of unique values.
  • Each value of inorder also appears in preorder.
  • preorder is guaranteed to be the preorder traversal of the tree.
  • inorder is guaranteed to be the inorder traversal of the tree.

Solutions

Python3

# 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, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        if not preorder:
            return None
        v = preorder[0]
        root = TreeNode(val=v)
        i = inorder.index(v)
        root.left = self.buildTree(preorder[1 : 1 + i], inorder[:i])
        root.right = self.buildTree(preorder[1 + i :], inorder[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, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        def dfs(i, j, n):
            if n <= 0:
                return None
            v = preorder[i]
            k = d[v]
            root = TreeNode(v)
            root.left = dfs(i + 1, j, k - j)
            root.right = dfs(i + 1 + k - j, k + 1, n - k + j - 1)
            return root

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

Java

/**
 * 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[] preorder, int[] inorder) {
        for (int i = 0; i < inorder.length; ++i) {
            indexes.put(inorder[i], i);
        }
        return dfs(preorder, inorder, 0, 0, preorder.length);
    }

    private TreeNode dfs(int[] preorder, int[] inorder, int i, int j, int n) {
        if (n <= 0) {
            return null;
        }
        int v = preorder[i];
        int k = indexes.get(v);
        TreeNode root = new TreeNode(v);
        root.left = dfs(preorder, inorder, i + 1, j, k - j);
        root.right = dfs(preorder, inorder, i + 1 + k - j, k + 1, n - k + j - 1);
        return root;
    }
}

C++

/**
 * 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>& preorder, vector<int>& inorder) {
        for (int i = 0; i < inorder.size(); ++i) indexes[inorder[i]] = i;
        return dfs(preorder, inorder, 0, 0, inorder.size());
    }

    TreeNode* dfs(vector<int>& preorder, vector<int>& inorder, int i, int j, int n) {
        if (n <= 0) return nullptr;
        int v = preorder[i];
        int k = indexes[v];
        TreeNode* root = new TreeNode(v);
        root->left = dfs(preorder, inorder, i + 1, j, k - j);
        root->right = dfs(preorder, inorder, i + 1 + k - j, k + 1, n - k + j - 1);
        return root;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func buildTree(preorder []int, inorder []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 := preorder[i]
		k := indexes[v]
		root := &TreeNode{Val: v}
		root.Left = dfs(i+1, j, k-j)
		root.Right = dfs(i+1+k-j, k+1, n-k+j-1)
		return root
	}
	return dfs(0, 0, len(inorder))
}

TypeScript

/**
 * 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(preorder: number[], inorder: number[]): TreeNode | null {
    const n = preorder.length;
    if (n === 0) {
        return null;
    }
    const val = preorder[0];
    const index = inorder.indexOf(val);
    return new TreeNode(
        val,
        buildTree(preorder.slice(1, index + 1), inorder.slice(0, index)),
        buildTree(preorder.slice(index + 1), inorder.slice(index + 1)),
    );
}

Rust

// 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 to_tree(preorder: &[i32], inorder: &[i32]) -> Option<Rc<RefCell<TreeNode>>> {
        if preorder.is_empty() {
            return None;
        }
        let val = preorder[0];
        let index = inorder.iter().position(|&v| v == val).unwrap();
        Some(Rc::new(RefCell::new(TreeNode {
            val,
            left: Self::to_tree(&preorder[1..index + 1], &inorder[..index]),
            right: Self::to_tree(&preorder[index + 1..], &inorder[index + 1..]),
        })))
    }

    pub fn build_tree(preorder: Vec<i32>, inorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
        Self::to_tree(&preorder[..], &inorder[..])
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {number[]} preorder
 * @param {number[]} inorder
 * @return {TreeNode}
 */
var buildTree = function (preorder, inorder) {
    function dfs(i, j, n) {
        if (n <= 0) {
            return null;
        }
        const v = preorder[i];
        const k = d[v];
        const root = new TreeNode(v);
        root.left = dfs(i + 1, j, k - j);
        root.right = dfs(i + 1 + k - j, k + 1, n - k + j - 1);
        return root;
    }
    const d = new Map();
    for (const [i, v] of inorder.entries()) {
        d[v] = i;
    }
    return dfs(0, 0, inorder.length);
};

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