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102. Binary Tree Level Order Traversal

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:

img

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

Example 2:

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

Example 3:

Input: root = []
Output: []

Solution:

Tree里头关于层有关的请优先想到BFS, queue

/**
 * 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>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();

        int index = 0;

        dfs(root, index, result);
        return result;
    }

    private void dfs(TreeNode root, int index, List<List<Integer>> result){
        if (root == null){
            return;
        }

        if (index >= result.size()){
            result.add(new ArrayList<Integer>());
        }


        result.get(index).add(root.val);

        dfs(root.left, index + 1, result);
        dfs(root.right, index + 1, result);
    }
}

// TC: O(n)
// SC: O(n)

/**
 * 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>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (root == null){
            return result;
        }

        int index = 0;
        dfs(root, index, result);

        return result;
    }

    private void dfs (TreeNode root, int index, List<List<Integer>> result){
        if (root == null){
            return;
        }

        if (index >= result.size()){
            result.add(new ArrayList<Integer>());
        }

        result.get(index).add(root.val);
        dfs(root.left, index + 1, result);
        dfs(root.right, index + 1, result);
    }
}
// TC: O(n)
// SC: O(n)

/**
 * 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>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (root == null){
            return result;
        }

        int index = 0;
        DFS(root, index, result);

        return result;

    }

    private void DFS(TreeNode root, int index, List<List<Integer>> result){
        // base case
        if (root == null){
            return;
        }

        if (index >= result.size()){
            result.add(new ArrayList<Integer>());
        }

        result.get(index).add(root.val);
        DFS(root.left, index + 1, result);
        DFS(root.right, index +1 , result);
    }
}

// TC: O(n)
// SC: O(n)

Tree的题 代码答案不难理解, 单纯让你想思路 感觉很容易无从下手的感觉 -> 但基本上都要遍历