Leetcode/各大家 -- 102.107. Binary Tree Level Order Traversal II(BFS)

102.107. Binary Tree Level Order Traversal I&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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

102. return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

107. return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

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思路： BFS Using Queue, key is to track current level by queue.size()
Complexity: O(N) visit every node only once.

public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>> ();
        if(root == null)return res;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        while(!queue.isEmpty()){
            int size = queue.size();
            ArrayList<Integer> level = new ArrayList<Integer>();
            for (int i = 0; i < size; i++){
                TreeNode curr = queue.remove();
                level.add(curr.val);
                if(curr.left != null) queue.add(curr.left);
                if(curr.right != null) queue.add(curr.right);
            }
            res.add(level);//res.add(0,level);--bottom up
        }
        return res;
    }