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32: Binary Tree Level Order Traversal
/************************************************************************/
/* 32: Binary Tree Level Order Traversal */
/************************************************************************/
/*
* Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
* */
//从根节点到叶子节点分层遍历树
public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> result = new ArrayList<>(); List<TreeNode> level = new ArrayList<>(); level.add(root); while(true){ if (level.isEmpty() || level.get(0) == null){ break; } List<TreeNode> nextLevel = new ArrayList<>(); List<Integer> currentLevel = new ArrayList<>(); for (TreeNode node : level){ currentLevel.add(node.val); if (node.left != null) nextLevel.add(node.left); if (node.right != null) nextLevel.add(node.right); } result.add(currentLevel); level = nextLevel; } return result; }
32: Binary Tree Level Order Traversal
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