Binary Tree Level Order Traversal I & II

Given a binary tree, return thelevel ordertraversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree[3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

Note:

in Binary Tree Level Order Traversal II, the requirement is only different in getting the outcome as reverse order, namely bottom-up level order traversal as:

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

Analysis

Recursive - DFS

关键在传入一个结果的list,因为list中元素的序号代表了层数(n+1)

Queue

利用一个Queue记录需要扫描的层(level)的节点数(number of nodes),依次存入每个节点的左子节点和右子节点到Queue中,读取时依次从Queue取出节点并存入该层sub list,每层扫描完后存入总的wrap list。

Solution

Top-down level order traversal

Using Recursive - DFS
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        levelOrderHelper(root, 0, result);
        return result;
    }
    void levelOrderHelper(TreeNode root, int depth, List<List<Integer>> result) {
        if (root == null) {
            return;
        }
        if (depth == result.size()) {
            result.add(new ArrayList<Integer>());
        }
        result.get(depth).add(root.val);
        levelOrderHelper(root.left, depth + 1, result);
        levelOrderHelper(root.right, depth + 1, result);

    }
}
Using Queue - BFS
public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();

        if(root == null) return wrapList;

        queue.offer(root);
        while(!queue.isEmpty()){
            int levelNum = queue.size();
            List<Integer> subList = new LinkedList<Integer>();
            for(int i=0; i<levelNum; i++) {
                if(queue.peek().left != null) queue.offer(queue.peek().left);
                if(queue.peek().right != null) queue.offer(queue.peek().right);
                subList.add(queue.poll().val);
            }
            wrapList.add(subList);
        }
        return wrapList;
    }
}

Bottom-up level order traversal

DFS - Recursive
public class Solution {
    public List < List < Integer >> levelOrderBottom(TreeNode root) {
        List < List < Integer >> wrapList = new LinkedList < List < Integer >> ();
        levelMaker(wrapList, root, 0);
        return wrapList;
    }

    public void levelMaker(List < List < Integer >> list, TreeNode root, int level) {
        if (root == null) return;
        if (level >= list.size()) {
            list.add(0, new LinkedList < Integer > ());
        }
        levelMaker(list, root.left, level + 1);
        levelMaker(list, root.right, level + 1);
        list.get(list.size() - level - 1).add(root.val);
    }
}
BFS - Queue
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();

        if(root == null) return wrapList;

        queue.offer(root);
        while(!queue.isEmpty()){
            int levelNum = queue.size();
            List<Integer> subList = new LinkedList<Integer>();
            for(int i=0; i<levelNum; i++) {
                if(queue.peek().left != null) queue.offer(queue.peek().left);
                if(queue.peek().right != null) queue.offer(queue.peek().right);
                subList.add(queue.poll().val);
            }
            wrapList.add(0, subList);
        }
        return wrapList;
    }
}

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