107. Binary Tree Level Order Traversal II

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LeetCode

Tree, BFS

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/**
 * 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) {
        List<List<Integer>> res=new ArrayList<>();
        if(root==null) return res;
        Queue<TreeNode> q=new LinkedList<>();
        q.offer(root);
        while(!q.isEmpty()){
            List<Integer> tmp=new ArrayList<>();
            int qSize=q.size();
            while(qSize>0){
                TreeNode tmpNode=q.poll();
                tmp.add(tmpNode.val);
                if(tmpNode.left!=null) q.offer(tmpNode.left);
                if(tmpNode.right!=null) q.offer(tmpNode.right);
                qSize--;
            }
            res.add(0,tmp);
        }
        return res;
    }
}

DFS

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class Solution {
    List<List<Integer>> levels = new ArrayList<List<Integer>>();

    public void util(TreeNode node, int level) {
        // start the current level
        if (levels.size() == level) levels.add(new ArrayList<Integer>());

         // append the current node value
         levels.get(level).add(node.val);

         // process child nodes for the next level
         if (node.left != null) util(node.left, level + 1);
         if (node.right != null) util(node.right, level + 1);
    }
    
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        if (root == null) return levels;
        util(root, 0);
        Collections.reverse(levels);
        return levels;
    }
}

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