Binary Tree Level Order Traversal

最新推荐文章于 2022-02-20 23:20:19 发布

smallfish_yy

最新推荐文章于 2022-02-20 23:20:19 发布

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分类专栏： Binary Tree LeetCode tree traversal

本文链接：https://blog.youkuaiyun.com/smallfish_yy/article/details/17627001

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tree traversal

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本文介绍了二叉树的层次遍历算法，包括递归实现和广度优先搜索(BFS)两种方法，并提供了详细的代码示例。

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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},

return its level order traversal as:

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

Update 17/01/2014:

1. Recursive implementation

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void levelOrder(TreeNode root, ArrayList<ArrayList<Integer>> res, int level) {
        if(root==null) return;
        if(res.size()==level) res.add(new ArrayList<Integer>());
        res.get(level).add(root.val);
        levelOrder(root.left, res, level+1);
        levelOrder(root.right, res, level+1);
        return;
    }
    
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        levelOrder(root, res, 0);
        return res;
    }
}

2. BFS.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        Queue<TreeNode> frontier = new LinkedList<TreeNode>();
        if(root==null) return res;
        frontier.add(root);
        while(!frontier.isEmpty()) {
            Queue<TreeNode> next = new LinkedList<TreeNode>();
            ArrayList<Integer> tem = new ArrayList<Integer>();
            while(!frontier.isEmpty()) {
                TreeNode cur = frontier.poll();
                if(cur.left!=null) next.add(cur.left);
                if(cur.right!=null) next.add(cur.right);
                tem.add(cur.val);
            }
            res.add(tem);
            frontier = next;
        }
        return res;
    }
}

Recursive implementation:

Each node stores its own value into the res ArrayList first and then merge its left and right subtrees' results into it.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        ArrayList<ArrayList<Integer>> lres = new ArrayList<ArrayList<Integer>>();
        ArrayList<ArrayList<Integer>> rres = new ArrayList<ArrayList<Integer>>();
        ArrayList<Integer> temp = new ArrayList<Integer>();
        
        if(root==null) return res;
        
        temp.add(root.val);
        res.add(temp);
        if(root.left==null && root.right==null) return res;
        if(root.left != null) lres = levelOrder(root.left);
        if(root.right != null) rres = levelOrder(root.right);
        int i=0;
        while(i<lres.size() || i<rres.size()) {
            if(i >= lres.size()) res.add(rres.get(i));
            else if(i >= rres.size()) res.add(lres.get(i));
            else {
                lres.get(i).addAll(rres.get(i));
                res.add(lres.get(i));
            }
            i++;
        }
        return res;
    }
}