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;
}
}