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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

The approach is the same as in 102. Binary Tree Level Order Traversal, just reverse the result in the end.

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

  • /**
 * 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 i = q.size(); i; --i) {
                auto node = q.front();
                q.pop();
                t.emplace_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            ans.emplace_back(t);
        }
        reverse(ans.begin(), ans.end());
        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 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.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrderBottom(root *TreeNode) [][]int {
	ans := [][]int{}
	if root == nil {
		return ans
	}
	q := []*TreeNode{root}
	for len(q) > 0 {
		var t []int
		for i := len(q); i > 0; i-- {
			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([][]int{t}, ans...)
	}
	return 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 = [];
        for (let i = q.length; i > 0; --i) {
            const node = q.shift();
            t.push(node.val);
            if (node.left) q.push(node.left);
            if (node.right) q.push(node.right);
        }
        ans.unshift(t);
    }
    return ans;
};


  • // 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, cell::RefCell, collections::VecDeque };
impl Solution {
    #[allow(dead_code)]
    pub fn level_order_bottom(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
        if root.is_none() {
            return vec![];
        }
        let mut ret_vec = Vec::new();
        let mut q = VecDeque::new();

        q.push_back(root);

        while !q.is_empty() {
            let mut cur_vec = Vec::new();
            let mut next_q = VecDeque::new();
            while !q.is_empty() {
                let cur_front = q.front().unwrap().clone();
                q.pop_front();
                cur_vec.push(cur_front.as_ref().unwrap().borrow().val);
                let left = cur_front.as_ref().unwrap().borrow().left.clone();
                let right = cur_front.as_ref().unwrap().borrow().right.clone();
                if !left.is_none() {
                    next_q.push_back(left);
                }
                if !right.is_none() {
                    next_q.push_back(right);
                }
            }
            ret_vec.push(cur_vec);
            q = next_q;
        }

        ret_vec.reverse();
        ret_vec
    }
}


  • /**
 * 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();
}

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