Binary Tree Level Order Traversal II
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] ]
Naive Way: 这题，不就是之前那题把list的顺序倒过来吗。然后用了一个stack倒转顺序。
/**
* Definition for binary tree
* 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>> rlst = new ArrayList<List<Integer>>();
Stack<List<Integer>> stack = new Stack<List<Integer>>();
Map<TreeNode, Integer> map = new HashMap<TreeNode, Integer>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
if(root==null){return rlst;}
queue.add(root);
map.put(root,0);
while(!queue.isEmpty()){
TreeNode node = queue.poll();
int layer = map.get(node);
List<Integer> list;
if(rlst.size() > layer){
list = rlst.get(layer);
}else{
list = new ArrayList<Integer>();
rlst.add(list);
}
rlst.get(layer).add(node.val);
if(node.left!=null){
queue.add(node.left);
map.put(node.left, layer+1);
}
if(node.right!=null){
queue.add(node.right);
map.put(node.right, layer+1);
}
}
for(int i = 0;i < rlst.size();i++)
stack.push(rlst.get(i));
rlst.clear();
while(!stack.isEmpty()){
rlst.add(stack.pop());
}
return rlst;
}
}
Improved Way:能否直接得到倒转的list而不是得到正的再颠倒它呢。
