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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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