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:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
Map<Integer, List<Integer>> map = new TreeMap<>((a,b) -> {
return b - a;
});
dfs(root, 0, map);
System.out.println(map);
List<List<Integer>> result = new ArrayList<>();
for (List<Integer> levelNodes: map.values()) {
result.add(levelNodes);
}
return result;
}
void dfs(TreeNode node, int level, Map<Integer, List<Integer>> map) {
if (node == null) {
return;
}
dfs(node.left, level + 1, map);
addLevelNode(map, node.val, level);
dfs(node.right, level + 1, map);
}
void addLevelNode(Map<Integer, List<Integer>> map, int val, int level) {
List<Integer> levelList = map.getOrDefault(level, new ArrayList<>());
levelList.add(val);
map.put(level, levelList);
}
}