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Formatted question description: https://leetcode.ca/all/1305.html

1305. All Elements in Two Binary Search Trees (Medium)

Given two binary search trees root1 and root2.

Return a list containing all the integers from both trees sorted in ascending order.

 

Example 1:

Input: root1 = [2,1,4], root2 = [1,0,3]
Output: [0,1,1,2,3,4]

Example 2:

Input: root1 = [0,-10,10], root2 = [5,1,7,0,2]
Output: [-10,0,0,1,2,5,7,10]

Example 3:

Input: root1 = [], root2 = [5,1,7,0,2]
Output: [0,1,2,5,7]

Example 4:

Input: root1 = [0,-10,10], root2 = []
Output: [-10,0,10]

Example 5:

Input: root1 = [1,null,8], root2 = [8,1]
Output: [1,1,8,8]

 

Constraints:

  • Each tree has at most 5000 nodes.
  • Each node's value is between [-10^5, 10^5].

Related Topics:
Sort, Tree

Solution 1.

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
            List<Integer> elements1 = inorderTraversal(root1);
            List<Integer> elements2 = inorderTraversal(root2);
            List<Integer> allElements = new ArrayList<Integer>();
            int size1 = elements1.size(), size2 = elements2.size();
            int index1 = 0, index2 = 0;
            while (index1 < size1 && index2 < size2) {
                int element1 = elements1.get(index1), element2 = elements2.get(index2);
                if (element1 <= element2) {
                    allElements.add(element1);
                    index1++;
                } else {
                    allElements.add(element2);
                    index2++;
                }
            }
            while (index1 < size1) {
                int element1 = elements1.get(index1);
                allElements.add(element1);
                index1++;
            }
            while (index2 < size2) {
                int element2 = elements2.get(index2);
                allElements.add(element2);
                index2++;
            }
            return allElements;
        }
    
        public List<Integer> inorderTraversal(TreeNode root) {
            List<Integer> inorderTraversal = new ArrayList<Integer>();
            Stack<TreeNode> stack = new Stack<TreeNode>();
            TreeNode node = root;
            while (!stack.isEmpty() || node != null) {
                while (node != null) {
                    stack.push(node);
                    node = node.left;
                }
                TreeNode visitNode = stack.pop();
                inorderTraversal.add(visitNode.val);
                node = visitNode.right;
            }
            return inorderTraversal;
        }
    }
    
    ############
    
    /**
     * 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;
        }
    }
    
  • // OJ: https://leetcode.com/problems/all-elements-in-two-binary-search-trees/
    // Time: O(A + B)
    // Space: O(HA + HB)
    class BstIterator {
        stack<TreeNode*> s;
        void pushNodes(TreeNode *node) {
            for (; node; node = node->left) s.push(node);
        }
    public:
        BstIterator(TreeNode *root) {
            pushNodes(root);
        }
        int peek() {
            return hasNext() ? s.top()->val : INT_MIN;
        }
        bool hasNext() {
            return s.size();
        }
        void next() {
            if (!hasNext()) return;
            auto n = s.top();
            s.pop();
            pushNodes(n->right);
        }
    };
    class Solution {
    public:
        vector<int> getAllElements(TreeNode* a, TreeNode* b) {
            BstIterator i(a), j(b);
            vector<int> ans;
            while (i.hasNext() || j.hasNext()) {
                if (!j.hasNext() || (i.hasNext() && i.peek() <= j.peek())) {
                    ans.push_back(i.peek());
                    i.next();
                } else {
                    ans.push_back(j.peek());
                    j.next();
                }
            }
            return ans;
        }
    };
    
  • # 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)
    
    ############
    
    # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, val=0, left=None, right=None):
    #         self.val = val
    #         self.left = left
    #         self.right = right
    class Solution(object):
        def getAllElements(self, root1, root2):
            """
            :type root1: TreeNode
            :type root2: TreeNode
            :rtype: List[int]
            """
            nums1 = []
            nums2 = []
            self.inOrder(root1, nums1)
            self.inOrder(root2, nums2)
            return self.merge(nums1, nums2)
            
            
        def inOrder(self, root, nums):
            if not root:
                return
            self.inOrder(root.left, nums)
            nums.append(root.val)
            self.inOrder(root.right, nums)
        
        def merge(self, nums1, nums2):
            res = []
            p1 = 0
            p2 = 0
            while p1 < len(nums1) and p2 < len(nums2):
                if nums1[p1] < nums2[p2]:
                    res.append(nums1[p1])
                    p1 += 1
                else:
                    res.append(nums2[p2])
                    p2 += 1
            if p1 < len(nums1):
                res.extend(nums1[p1:])
            if p2 < len(nums2):
                res.extend(nums2[p2:])
            return res
    
  • /**
     * 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)
    }
    
  • /**
     * 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;
    }
    
    
  • // 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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