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leetcode 107 Binary Tree Level Order Traversal II ----- java

2024-08-18 01:28:28   219人阅读

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

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   /   9  20
    /     15   7

 

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

与之前的题做法一样，就是插入的时候，每次在第一位插入。

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
            

        List list = new ArrayList<List<Integer>>();

        if( root == null)
            return list;

        Queue<TreeNode> tree = new LinkedList<TreeNode>();


        tree.add(root);
        while( !tree.isEmpty() ){
            List ans = new ArrayList<Integer>();
            int size = tree.size();
            for( int i = 0;i<size;i++){
                TreeNode node = tree.poll();
                ans.add(node.val);
                if( node.left != null )
                    tree.add(node.left);
                if( node.right != null)
                    tree.add(node.right);
            }
            if( !ans.isEmpty() )
                list.add(0,ans);



        }



        return list;


    }
}

 

 

 

 

 

leetcode 107 Binary Tree Level Order Traversal II ----- java


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