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1305. 两棵二叉搜索树中的所有元素

题目描述

给你 root1 和 root2 这两棵二叉搜索树。请你返回一个列表，其中包含 两棵树 中的所有整数并按 升序 排序。.

 

示例 1：

输入：root1 = [2,1,4], root2 = [1,0,3]
输出：[0,1,1,2,3,4]

示例 2：

输入：root1 = [1,null,8], root2 = [8,1]
输出：[1,1,8,8]

 

提示：

  • 每棵树的节点数在 [0, 5000] 范围内
  • -105 <= Node.val <= 105

解法

方法一

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# 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 getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
        def dfs(root, t):
            if root is None:
                return
            dfs(root.left, t)
            t.append(root.val)
            dfs(root.right, t)

        def merge(t1, t2):
            ans = []
            i = j = 0
            while i < len(t1) and j < len(t2):
                if t1[i] <= t2[j]:
                    ans.append(t1[i])
                    i += 1
                else:
                    ans.append(t2[j])
                    j += 1
            while i < len(t1):
                ans.append(t1[i])
                i += 1
            while j < len(t2):
                ans.append(t2[j])
                j += 1
            return ans

        t1, t2 = [], []
        dfs(root1, t1)
        dfs(root2, t2)
        return merge(t1, t2)

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/**
 * 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 {
    public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
        List<Integer> t1 = new ArrayList<>();
        List<Integer> t2 = new ArrayList<>();
        dfs(root1, t1);
        dfs(root2, t2);
        return merge(t1, t2);
    }

    private void dfs(TreeNode root, List<Integer> t) {
        if (root == null) {
            return;
        }
        dfs(root.left, t);
        t.add(root.val);
        dfs(root.right, t);
    }

    private List<Integer> merge(List<Integer> t1, List<Integer> t2) {
        List<Integer> ans = new ArrayList<>();
        int i = 0, j = 0;
        while (i < t1.size() && j < t2.size()) {
            if (t1.get(i) <= t2.get(j)) {
                ans.add(t1.get(i++));
            } else {
                ans.add(t2.get(j++));
            }
        }
        while (i < t1.size()) {
            ans.add(t1.get(i++));
        }
        while (j < t2.size()) {
            ans.add(t2.get(j++));
        }
        return ans;
    }
}

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/**
 * 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:
    vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
        vector<int> t1;
        vector<int> t2;
        dfs(root1, t1);
        dfs(root2, t2);
        return merge(t1, t2);
    }

    void dfs(TreeNode* root, vector<int>& t) {
        if (!root) return;
        dfs(root->left, t);
        t.push_back(root->val);
        dfs(root->right, t);
    }

    vector<int> merge(vector<int>& t1, vector<int>& t2) {
        vector<int> ans;
        int i = 0, j = 0;
        while (i < t1.size() && j < t2.size()) {
            if (t1[i] <= t2[j])
                ans.push_back(t1[i++]);
            else
                ans.push_back(t2[j++]);
        }
        while (i < t1.size()) ans.push_back(t1[i++]);
        while (j < t2.size()) ans.push_back(t2[j++]);
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func getAllElements(root1 *TreeNode, root2 *TreeNode) []int {
    var dfs func(root *TreeNode) []int
    dfs = func(root *TreeNode) []int {
        if root == nil {
            return []int{}
        }
        left := dfs(root.Left)
        right := dfs(root.Right)
        left = append(left, root.Val)
        left = append(left, right...)
        return left
    }
    merge := func(t1, t2 []int) []int {
        var ans []int
        i, j := 0, 0
        for i < len(t1) && j < len(t2) {
            if t1[i] <= t2[j] {
                ans = append(ans, t1[i])
                i++
            } else {
                ans = append(ans, t2[j])
                j++
            }
        }
        for i < len(t1) {
            ans = append(ans, t1[i])
            i++
        }
        for j < len(t2) {
            ans = append(ans, t2[j])
            j++
        }
        return ans
    }
    t1, t2 := dfs(root1), dfs(root2)
    return merge(t1, t2)
}

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/**
 * 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 getAllElements(root1: TreeNode | null, root2: TreeNode | null): number[] {
    const res = [];
    const stacks = [[], []];
    while (root1 != null || stacks[0].length !== 0 || root2 != null || stacks[1].length !== 0) {
        if (root1 != null) {
            stacks[0].push(root1);
            root1 = root1.left;
        } else if (root2 != null) {
            stacks[1].push(root2);
            root2 = root2.left;
        } else {
            if (
                (stacks[0][stacks[0].length - 1] ?? { val: Infinity }).val <
                (stacks[1][stacks[1].length - 1] ?? { val: Infinity }).val
            ) {
                const { val, right } = stacks[0].pop();
                res.push(val);
                root1 = right;
            } else {
                const { val, right } = stacks[1].pop();
                res.push(val);
                root2 = right;
            }
        }
    }
    return res;
}

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// 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::rc::Rc;
impl Solution {
    pub fn get_all_elements(
        root1: Option<Rc<RefCell<TreeNode>>>,
        root2: Option<Rc<RefCell<TreeNode>>>,
    ) -> Vec<i32> {
        fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, t: &mut Vec<i32>) {
            if let Some(root) = root {
                dfs(&root.borrow().left, t);
                t.push(root.borrow().val);
                dfs(&root.borrow().right, t);
            }
        }

        let mut t1 = Vec::new();
        let mut t2 = Vec::new();
        dfs(&root1, &mut t1);
        dfs(&root2, &mut t2);

        let mut ans = Vec::new();
        let mut i = 0;
        let mut j = 0;
        while i < t1.len() && j < t2.len() {
            if t1[i] < t2[j] {
                ans.push(t1[i]);
                i += 1;
            } else {
                ans.push(t2[j]);
                j += 1;
            }
        }
        while i < t1.len() {
            ans.push(t1[i]);
            i += 1;
        }
        while j < t2.len() {
            ans.push(t2[j]);
            j += 1;
        }
        ans
    }
}

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