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leetcode102. binary tree level order traversal

admin 11月 13, 2020 0

二叉树的层次遍历的递归和迭代算法。

102. 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,null,null,15,7],

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  3
 / 
9  20
  /  
 15   7

return its level order traversal as:

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

迭代算法：

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public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> ans = new ArrayList<>();
    if (root == null) return ans;
    Queue<TreeNode> queue = new ArrayDeque<>();
    queue.offer(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        List<Integer> tmp = new ArrayList<>();
        for (int i = 0; i < size; i++) {
            TreeNode p = queue.poll();
            tmp.add(p.val);
            if (p.left != null) queue.offer(p.left);
            if (p.right != null) queue.offer(p.right);
        }
        ans.add(tmp);
    }
    return ans;
}

迭代算法：

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public List<List<Integer>> levelOrder(TreeNode root) {
    List<List<Integer>> ans = new ArrayList<>();
    if (root == null) return ans;
    levelOrderWalk(root, 0, ans);
    return ans;
}
public void (TreeNode root, int level, List<List<Integer>> ans) {
    if (root == null) return;
    if (level >= ans.size()) ans.add(new ArrayList<>());
    ans.get(level).add(root.val);
    levelOrderWalk(root.left, level+1, ans);
    levelOrderWalk(root.right, level+1, ans);
}
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