• 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,#,#,15,7},
    3
    /
    9 20
    /
    15 7
    return its level order traversal as:
    [
    [3],
    [9,20],
    [15,7]
    ]

    DFS实现:

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

    BFS实现:

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        // BFS遍历打印
        public List<List<Integer>> levelOrder(TreeNode root) {
            List<List<Integer>> res = new LinkedList<List<Integer>>();
            if (root == null)
                return res;
            LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
            queue.add(root);
            ArrayList<Integer> list = new ArrayList<Integer>();
            int cur = 1;
            int nextLevel = 0;
            while (!queue.isEmpty()) {
                TreeNode tmp = queue.poll();
    
                cur--;
                if (tmp.left != null) {
                    queue.add(tmp.left);
                    nextLevel++;
                }
                if (tmp.right != null) {
                    queue.add(tmp.right);
                    nextLevel++;
                }
                list.add(tmp.val);
                if (cur == 0) {
                    res.add(list);
                    list = new ArrayList<Integer>();
                    cur = nextLevel;
                    nextLevel = 0;
                }
    
            }
            return res;
        }
    
    }
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  • 原文地址:https://www.cnblogs.com/tina-smile/p/5056199.html
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