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105. 从前序与中序遍历序列构造二叉树

题目描述

给定两个整数数组 preorder 和 inorder ，其中 preorder 是二叉树的先序遍历， inorder 是同一棵树的中序遍历，请构造二叉树并返回其根节点。

示例 1:

输入: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
输出: [3,9,20,null,null,15,7]

示例 2:

输入: preorder = [-1], inorder = [-1]
输出: [-1]

提示:

1 <= preorder.length <= 3000
inorder.length == preorder.length
-3000 <= preorder[i], inorder[i] <= 3000
preorder 和 inorder 均 无重复 元素
inorder 均出现在 preorder
preorder 保证为二叉树的前序遍历序列
inorder 保证为二叉树的中序遍历序列

解法

方法一：哈希表 + 递归

前序序列的第一个节点 $preorder[0]$ 为根节点，我们在中序序列中找到根节点的位置 $k$，可以将中序序列划分为左子树 $inorder[0..k]$ 、右子树 $inorder[k+1..]$。

通过左右子树的区间，可以计算出左、右子树节点的个数，假设为 $a$ 和 $b$。然后在前序节点中，从根节点往后的 $a$ 个节点为左子树，再往后的 $b$ 个节点为右子树。

因此，我们设计一个函数 $dfs(i, j, n)$，其中 $i$ 和 $j$ 分别表示前序序列和中序序列的起始位置，而 $n$ 表示节点个数。函数的返回值是以 $preorder[i..i+n-1]$ 为前序序列，以 $inorder[j..j+n-1]$ 为中序序列构造出的二叉树。

函数 $dfs(i, j, n)$ 的执行过程如下：

如果 $n \leq 0$，说明没有节点，返回空节点。
取出前序序列的第一个节点 $v = preorder[i]$ 作为根节点，然后利用哈希表 $d$ 找到根节点在中序序列中的位置 $k$，那么左子树的节点个数为 $k - j$，右子树的节点个数为 $n - k + j - 1$。
递归构造左子树 $l = dfs(i + 1, j, k - j)$ 和右子树 $r = dfs(i + 1 + k - j, k + 1, n - k + j - 1)$。
最后返回以 $v$ 为根节点且左右子树分别为 $l$ 和 $r$ 的二叉树。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树节点个数。

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]:
        def dfs(i: int, j: int, n: int) -> Optional[TreeNode]:
            if n <= 0:
                return None
            v = preorder[i]
            k = d[v]
            l = dfs(i + 1, j, k - j)
            r = dfs(i + 1 + k - j, k + 1, n - k + j - 1)
            return TreeNode(v, l, r)

        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 int[] preorder;
    private Map<Integer, Integer> d = new HashMap<>();

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int n = preorder.length;
        this.preorder = preorder;
        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 = preorder[i];
        int k = d.get(v);
        TreeNode l = dfs(i + 1, j, k - j);
        TreeNode r = dfs(i + 1 + k - j, k + 1, n - 1 - (k - j));
        return new TreeNode(v, l, r);
    }
}

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

Go

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

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 d: Map<number, number> = new Map();
    const n = inorder.length;
    for (let i = 0; i < n; ++i) {
        d.set(inorder[i], i);
    }
    const dfs = (i: number, j: number, n: number): TreeNode | null => {
        if (n <= 0) {
            return null;
        }
        const v = preorder[i];
        const k = d.get(v)!;
        const l = dfs(i + 1, j, k - j);
        const r = dfs(i + 1 + k - j, k + 1, n - 1 - (k - j));
        return new TreeNode(v, l, r);
    };
    return dfs(0, 0, n);
}

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::cell::RefCell;
use std::collections::HashMap;
use std::rc::Rc;
impl Solution {
    pub fn build_tree(preorder: Vec<i32>, inorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
        let mut d = HashMap::new();
        for (i, &x) in inorder.iter().enumerate() {
            d.insert(x, i);
        }
        Self::dfs(&preorder, &d, 0, 0, preorder.len())
    }

    pub fn dfs(
        preorder: &Vec<i32>,
        d: &HashMap<i32, usize>,
        i: usize,
        j: usize,
        n: usize,
    ) -> Option<Rc<RefCell<TreeNode>>> {
        if n <= 0 {
            return None;
        }
        let v = preorder[i];
        let k = d[&v];
        let mut root = TreeNode::new(v);
        root.left = Self::dfs(preorder, d, i + 1, j, k - j);
        root.right = Self::dfs(preorder, d, i + k - j + 1, k + 1, n - k + j - 1);
        Some(Rc::new(RefCell::new(root)))
    }
}

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) {
    const d = new Map();
    const n = inorder.length;
    for (let i = 0; i < n; ++i) {
        d.set(inorder[i], i);
    }
    const dfs = (i, j, n) => {
        if (n <= 0) {
            return null;
        }
        const v = preorder[i];
        const k = d.get(v);
        const l = dfs(i + 1, j, k - j);
        const r = dfs(i + 1 + k - j, k + 1, n - 1 - (k - j));
        return new TreeNode(v, l, r);
    };
    return dfs(0, 0, n);
};

如果题目中给定的节点值存在重复，那么我们只需要记录每个节点值出现的所有位置，然后递归构建出所有可能的二叉树即可。

Python3

class Solution:
    def getBinaryTrees(self, preOrder: List[int], inOrder: List[int]) -> List[TreeNode]:
        def dfs(i: int, j: int, n: int) -> List[TreeNode]:
            if n <= 0:
                return [None]
            v = preOrder[i]
            ans = []
            for k in d[v]:
                if j <= k < j + n:
                    for l in dfs(i + 1, j, k - j):
                        for r in dfs(i + 1 + k - j, k + 1, n - 1 - (k - j)):
                            ans.append(TreeNode(v, l, r))
            return ans

        d = defaultdict(list)
        for i, x in enumerate(inOrder):
            d[x].append(i)
        return dfs(0, 0, len(preOrder))

Java

class Solution {
    private List<Integer> preorder;
    private Map<Integer, List<Integer>> d = new HashMap<>();

    public List<TreeNode> getBinaryTrees(List<Integer> preOrder, List<Integer> inOrder) {
        int n = preOrder.size();
        this.preorder = preOrder;
        for (int i = 0; i < n; ++i) {
            d.computeIfAbsent(inOrder.get(i), k -> new ArrayList<>()).add(i);
        }
        return dfs(0, 0, n);
    }

    private List<TreeNode> dfs(int i, int j, int n) {
        List<TreeNode> ans = new ArrayList<>();
        if (n <= 0) {
            ans.add(null);
            return ans;
        }
        int v = preorder.get(i);
        for (int k : d.get(v)) {
            if (k >= j && k < j + n) {
                for (TreeNode l : dfs(i + 1, j, k - j)) {
                    for (TreeNode r : dfs(i + 1 + k - j, k + 1, n - 1 - (k - j))) {
                        ans.add(new TreeNode(v, l, r));
                    }
                }
            }
        }
        return ans;
    }
}

C++

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> getBinaryTrees(vector<int>& preOrder, vector<int>& inOrder) {
        int n = inOrder.size();
        unordered_map<int, vector<int>> d;
        for (int i = 0; i < n; ++i) {
            d[inOrder[i]].push_back(i);
        }
        function<vector<TreeNode*>(int, int, int)> dfs = [&](int i, int j, int n) -> vector<TreeNode*> {
            vector<TreeNode*> ans;
            if (n <= 0) {
                ans.push_back(nullptr);
                return ans;
            }
            int v = preOrder[i];
            for (int k : d[v]) {
                if (k >= j && k < j + n) {
                    auto lefts = dfs(i + 1, j, k - j);
                    auto rights = dfs(i + 1 + k - j, k + 1, n - 1 - (k - j));
                    for (TreeNode* l : lefts) {
                        for (TreeNode* r : rights) {
                            TreeNode* node = new TreeNode(v);
                            node->left = l;
                            node->right = r;
                            ans.push_back(node);
                        }
                    }
                }
            }
            return ans;
        };
        return dfs(0, 0, n);
    }
};

Go

func getBinaryTrees(preOrder []int, inOrder []int) []*TreeNode {
	n := len(preOrder)
	d := map[int][]int{}
	for i, x := range inOrder {
		d[x] = append(d[x], i)
	}
	var dfs func(i, j, n int) []*TreeNode
	dfs = func(i, j, n int) []*TreeNode {
		ans := []*TreeNode{}
		if n <= 0 {
			ans = append(ans, nil)
			return ans
		}
		v := preOrder[i]
		for _, k := range d[v] {
			if k >= j && k < j+n {
				lefts := dfs(i+1, j, k-j)
				rights := dfs(i+1+k-j, k+1, n-1-(k-j))
				for _, left := range lefts {
					for _, right := range rights {
						ans = append(ans, &TreeNode{v, left, right})
					}
				}
			}
		}
		return ans
	}
	return dfs(0, 0, n)
}

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105. 从前序与中序遍历序列构造二叉树

题目描述

解法

方法一：哈希表 + 递归

Python3

Java

C++

Go

TypeScript

Rust

JavaScript

Python3

Java

C++

Go

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105. 从前序与中序遍历序列构造二叉树

题目描述

解法

方法一：哈希表 + 递归

Python3

Java

C++

Go

TypeScript

Rust

JavaScript

Python3

Java

C++

Go