LeetCode 102. Binary Tree Level Order Traversal

Question

Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).

Example 1:

tree1
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Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]

Example 2:

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Input: root = [1]
Output: [[1]]

Example 3:

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Input: root = []
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 2000].
  • -1000 <= Node.val <= 1000

Source: https://leetcode.com/problems/binary-tree-level-order-traversal/

Solution

Use queue size in BFS to implement layer-ordre traversal of a tree.

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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;
}
}

public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
if (root != null) {
queue.add(root);
}
while (!queue.isEmpty()) {
List<Integer> currLevel = new ArrayList<>();
int numThisLevel = queue.size();
for (int i = 0; i < numThisLevel; i++) {
TreeNode curr = queue.poll();
currLevel.add(curr.val);
if (curr.left != null) {
queue.add(curr.left);
}
if (curr.right != null) {
queue.add(curr.right);
}
}
result.add(currLevel);
}
return result;
}
Author

Weihao Ye

Posted on

2022-01-25

Updated on

2022-01-25

Licensed under