107. Binary Tree Level Order Traversal II
Tags: ‘Tree’, ‘Breadth-first Search’
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] ]
Solution
类比102 方法1：层级遍历 然后reverse
public List<List<Integer>> levelOrderBottom(TreeNode root) {
LinkedList<List<Integer>> result = new LinkedList<>();
if (root == null) return result;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
List<Integer> list = new LinkedList<>();
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
list.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
result.addFirst(list);
// or user Collections.reverse(result)
}
return result;
}
方法2: DFS Recursive
public List<List<Integer>> levelOrderBottom(TreeNode root) {
LinkedList<List<Integer>> result = new LinkedList<List<Integer>>();
helper(root, result, 0);
return result;
}
private void helper(TreeNode root, LinkedList<List<Integer>> result, int height) {
if (root == null) return;
if (height >= result.size()) {
result.addFirst(new LinkedList<Integer>());
}
result.get(result.size() - 1 - height).add(root.val);
helper(root.left, result, height + 1);
helper(root.right, result, height + 1);
}