LeetCode Binary Tree Level Order Traversal II

最新推荐文章于 2021-11-18 16:57:58 发布

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本文介绍了一种从叶节点到根节点的层次遍历二叉树的方法，并提供了一个具体的实现示例。通过使用队列来跟踪每个层级的节点，最终得到一个倒序的层次遍历结果。

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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},

return its bottom-up level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

题意：倒着层次遍历一棵树。

思路：还是一样，最后倒置一下就行了

/**
 * 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>> ans = new ArrayList<List<Integer>>();
        if (root == null) return ans;
        
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.add(root);
        int count = 1, level = 0;
        while (!q.isEmpty()) {
        	List<Integer> tmp = new ArrayList<Integer>();
        	level = 0;
        	for (int i = 0; i < count; i++)  {
        		TreeNode t = q.poll();
        		tmp.add(t.val);
        		if (t.left != null) {
        			q.add(t.left);
        			++level;
        		}
        		if (t.right != null) {
        			q.add(t.right);
        			++level;
        		}
        	}
        	
        	count = level;
        	ans.add(tmp);
        }
        
        Collections.reverse(ans);
        return ans;
    }
}