[LintCode/LeetCode] Binary Tree Level Order Traver

makeFoxPlay 发布于2019-08-14 15:48 / 3027人阅读

Problem

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]
]

Note

用先入先出的Queue。用size标记每一层的结点数目并定向循环。

Solution

class Solution {
    public List> levelOrder(TreeNode root) {
        List> res = new ArrayList<>();
        if (root == null) return res;
        Queue queue = new LinkedList();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List temp = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
                temp.add(cur.val);
            }
            res.add(temp);//for traversal II: res.add(0, temp);
        }
        return res;
    }
}

DFS

class Solution {
    public List> levelOrder(TreeNode root) {
        List> res = new ArrayList<>();
        dfs(root, 0, res);
        return res;
    }
    private void dfs(TreeNode root, int level, List> res) {
        if (root == null) return;
        if (level >= res.size()) res.add(new ArrayList()); //level >= res.size() ---- from 0 >= 0 to initialize
        res.get(level).add(root.val);
        dfs(root.left, level+1, res);
        dfs(root.right, level+1, res);
    }
}

From bottom

public class Solution {
    public ArrayList> levelOrderBottom(TreeNode root) {
        ArrayList> res = new ArrayList<>();
        if (root == null) return res;
        helper(root, res, 0);
        return res;
    }
    public void helper(TreeNode root, ArrayList> res, int index) {
        int size = res.size();
        if (index == size) {
            ArrayList curList = new ArrayList<>();
            curList.add(root.val);
            res.add(0, curList);
        }
        else res.get(size-index-1).add(root.val);
        if (root.left != null) helper(root.left, res, index+1);
        if (root.right != null) helper(root.right, res, index+1);
    }
}

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