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algorithm notes: leetcode#107 binary tree level order traversal ii

admin 11月 12, 2020 0

Problem

Solution

Initial thoughts

Python implementation

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class :
    def levelOrderBottom(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        def dfs(node, depth):
            if node is None:
                return
            if len(self.ans) > depth:
                self.ans[-1-depth].append(node.val)
            else:
                self.ans = [[node.val]]+self.ans
            dfs(node.left, depth+1)
            dfs(node.right, depth+1)
        
        self.ans = []
        dfs(root, 0)
        return self.ans

Java implementation

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class  {
    private List<List<Integer>> ans = new ArrayList<>();
    private void dfs(TreeNode node, int depth){
        if(node==null){ return ; }
        if(ans.size() > depth){ ans.get(ans.size()-1-depth).add(node.val); }
        else{ 
            List<Integer> temp = new ArrayList<>();
            temp.add(node.val);
            ans.add(0, temp);
        }
        dfs(node.left, depth+1);
        dfs(node.right, depth+1);
    }
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        dfs(root, 0);
        return ans;
    }
}

Time complexity

O(n)

Space complexity

O(n)

Links

107. Binary Tree Level Order Traversal II
(中文版) 算法笔记: 力扣#107 二叉树的层次遍历 II