107. Binary Tree Level Order Traversal II 遍历二叉树2

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]
]

题意：用广度优先的方式，遍历二叉树

解法：借助队列存储节点，提前保存每一层的节点数量

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<int>> LevelOrderBottom(TreeNode root) {
		List<List<int>> result = new List<List<int>>();
		if (root == null) return result;
		Queue<TreeNode> queue = new Queue<TreeNode>();
		queue.Enqueue(root);
		while (queue.Count > 0) {
			List<int> curLevel = new List<int>();
			int count = queue.Count;
			for (int i = 0; i < count; i++) {
				TreeNode curNode = queue.Dequeue();
				curLevel.Add(curNode.val);
				if (curNode.left != null) queue.Enqueue(curNode.left);
				if (curNode.right != null) queue.Enqueue(curNode.right);
			}
			result.Add(curLevel);
		}
		result.Reverse();
		return result;
    }
}

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