Level Order Traversal II

出处

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

Solution

Queue version. On the basis of 'Binary Tree Level Order Traversal', reverse the final vector.

Complexity

时间复杂度 O(n),空间复杂度 O(n)

Code

/**
 * 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<Integer>> res = new ArrayList<List<Integer>>();
        if (root == null) return res;
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(root);
        q.offer(null);
        List<Integer> level = new ArrayList<Integer>();
        
        while(true) {
            TreeNode node = q.poll();
            if (node != null) {
                level.add(node.val);
                if(node.left!=null) q.offer(node.left);
                if(node.right!=null) q.offer(node.right);
            } else {
                res.add(level);
                level = new ArrayList<Integer>();
                if(q.isEmpty()==true) break;
                q.offer(null);
            }
        }
        Collections.reverse(res);
        return res;
    }
}