LeetCode——Binary Tree Level Order Traversal II

标签：leetcode

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]
]

题目：给定一个二叉树，从底到顶返回节点的层序遍历的值。（如，从左到右，一层一层）

思路：上一题的结果上面逆转一下即可。完全不用出这题吧？

	public List<List<Integer>> levelOrderBottom(TreeNode root) {
		List<List<Integer>> list = new ArrayList<List<Integer>>();
		if (root == null)
			return list;
		Queue<TreeNode> queue = new LinkedList<TreeNode>();
		queue.add(root);
		while (!queue.isEmpty()) {
			List<Integer> li = new ArrayList<Integer>();
			int size = queue.size();
			for (int i = 0; i < size; i++) {
				TreeNode node = queue.poll();
				li.add(node.val);
				if (node.left != null)
					queue.add(node.left);
				if (node.right != null)
					queue.add(node.right);
			}
			list.add(li);
		}
		List<List<Integer>> ret = new ArrayList<List<Integer>>();
		for (int i = list.size() - 1; i >= 0; i--) {
			ret.add(list.get(i));
		}
		return ret;
	}

	// Definition for binary tree
	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}

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LeetCode——Binary Tree Level Order Traversal II

标签：leetcode

原文地址：http://blog.csdn.net/laozhaokun/article/details/37757409