107. Binary Tree Level Order Traversal II

Description

Given the root of a binary tree, return the bottom-up level order traversal of its nodes' values. (i.e., from left to right, level by level from leaf to root).

Example 1:

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

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

Solutions

Solution 1: BFS

We can use the BFS (Breadth-First Search) method to solve this problem. First, enqueue the root node, then continuously perform the following operations until the queue is empty:

Traverse all nodes in the current queue, store their values in a temporary array \(t\), and then enqueue their child nodes.
Store the temporary array \(t\) in the answer array.

Finally, return the reversed answer array.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the number of nodes in the binary tree.

Python3JavaC++GoTypeScriptRustJavaScript

# 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 levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q)):
                node = q.popleft()
                t.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(t)
        return ans[::-1]

/**
 * 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<List<Integer>> levelOrderBottom(TreeNode root) {
        LinkedList<List<Integer>> ans = new LinkedList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> q = new LinkedList<>();
        q.offerLast(root);
        while (!q.isEmpty()) {
            List<Integer> t = new ArrayList<>();
            for (int i = q.size(); i > 0; --i) {
                TreeNode node = q.pollFirst();
                t.add(node.val);
                if (node.left != null) {
                    q.offerLast(node.left);
                }
                if (node.right != null) {
                    q.offerLast(node.right);
                }
            }
            ans.addFirst(t);
        }
        return ans;
    }
}

/**
 * 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<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> ans;
        if (!root) {
            return ans;
        }
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            vector<int> t;
            for (int n = q.size(); n; --n) {
                auto node = q.front();
                q.pop();
                t.push_back(node->val);
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
            ans.push_back(t);
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrderBottom(root *TreeNode) (ans [][]int) {
    if root == nil {
        return
    }
    q := []*TreeNode{root}
    for len(q) > 0 {
        t := []int{}
        for n := len(q); n > 0; n-- {
            node := q[0]
            q = q[1:]
            t = append(t, node.Val)
            if node.Left != nil {
                q = append(q, node.Left)
            }
            if node.Right != nil {
                q = append(q, node.Right)
            }
        }
        ans = append(ans, t)
    }
    for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 {
        ans[i], ans[j] = ans[j], ans[i]
    }
    return
}

/**
 * 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 levelOrderBottom(root: TreeNode | null): number[][] {
    const ans: number[][] = [];
    if (!root) {
        return ans;
    }
    const q: TreeNode[] = [root];
    while (q.length) {
        const t: number[] = [];
        const qq: TreeNode[] = [];
        for (const { val, left, right } of q) {
            t.push(val);
            left && qq.push(left);
            right && qq.push(right);
        }
        ans.push(t);
        q.splice(0, q.length, ...qq);
    }
    return ans.reverse();
}

// 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, collections::VecDeque, rc::Rc};
impl Solution {
    pub fn level_order_bottom(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
        let mut ans = Vec::new();
        if let Some(root_node) = root {
            let mut q = VecDeque::new();
            q.push_back(root_node);
            while !q.is_empty() {
                let mut t = Vec::new();
                for _ in 0..q.len() {
                    if let Some(node) = q.pop_front() {
                        let node_ref = node.borrow();
                        t.push(node_ref.val);
                        if let Some(ref left) = node_ref.left {
                            q.push_back(Rc::clone(left));
                        }
                        if let Some(ref right) = node_ref.right {
                            q.push_back(Rc::clone(right));
                        }
                    }
                }
                ans.push(t);
            }
        }
        ans.reverse();
        ans
    }
}

/**
 * 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 {TreeNode} root
 * @return {number[][]}
 */
var levelOrderBottom = function (root) {
    const ans = [];
    if (!root) {
        return ans;
    }
    const q = [root];
    while (q.length) {
        const t = [];
        const qq = [];
        for (const { val, left, right } of q) {
            t.push(val);
            left && qq.push(left);
            right && qq.push(right);
        }
        ans.push(t);
        q.splice(0, q.length, ...qq);
    }
    return ans.reverse();
};

107. Binary Tree Level Order Traversal II

Description

Solutions

Solution 1: BFS

Comments