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).
Solution
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
if(root == null)
return result;
queue.add(root);
while(queue.size()>0) {
List<Integer> list = new ArrayList<Integer>();
int size = queue.size();
for(int i=0; i<size; i++) {
TreeNode temp = queue.poll();
list.add(temp.val);
if(temp.left!=null) {
queue.add(temp.left);
}
if(temp.right!=null){
queue.add(temp.right);
}
}
result.add(0, list);
}
return result;
}
For example:
Given binary tree
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] ]
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
if(root == null)
return result;
queue.add(root);
while(queue.size()>0) {
List<Integer> list = new ArrayList<Integer>();
int size = queue.size();
for(int i=0; i<size; i++) {
TreeNode temp = queue.poll();
list.add(temp.val);
if(temp.left!=null) {
queue.add(temp.left);
}
if(temp.right!=null){
queue.add(temp.right);
}
}
result.add(0, list);
}
return result;
}
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