Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its bottom-up level order traversal as:

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

  解题思路和level order traversal 一样， 就是最后把输出结果反转一下

 一个queue的方法，reverse 结果

/**
 * 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>> levelOrderBottom(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        
        if(root == null)
            return result;
            
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        int nextLevelCount = 1;
        
        while(!queue.isEmpty()){
            int curLevel = nextLevelCount;
            nextLevelCount = 0;
            ArrayList<Integer> lvl = new ArrayList<Integer>();
            for(int i = 0; i < curLevel; i++){
                TreeNode tmp = queue.poll();
                lvl.add(tmp.val);
                if(tmp.left != null){
                    queue.add(tmp.left);
                    nextLevelCount++;
                }
                if(tmp.right != null){
                    queue.add(tmp.right);
                    nextLevelCount++;
                }
            }
            result.add(lvl);
        }
        ArrayList<ArrayList<Integer>> reverse = new ArrayList<ArrayList<Integer>>();
        for(int i = result.size()-1; i >= 0; i--){
            reverse.add(result.get(i));
        }
        return reverse;
    }
}

 用 Stack 和 两个queue 的方法

public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(root == null)
            return result;
        Stack<ArrayList<Integer>> stack = new Stack<ArrayList<Integer>>();
        stack = traversal(root);
        while(!stack.isEmpty())
            result.add(stack.pop());
        return result;
    }
    private Stack<ArrayList<Integer>> traversal(TreeNode root)
    {
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        Stack<ArrayList<Integer>> result = new Stack<ArrayList<Integer>>();
        queue.add(root);
        while(!queue.isEmpty())
        {
            ArrayList<Integer> tmp = new ArrayList<Integer>();
            LinkedList<TreeNode> tmpqueue = new LinkedList<TreeNode>();
            while(!queue.isEmpty())
            {
                TreeNode node = queue.poll();
                tmp.add(node.val);
                if(node.left != null)
                    tmpqueue.add(node.left);
                if(node.right != null)
                    tmpqueue.add(node.right);
            }
            result.add(tmp);
            queue = tmpqueue;
        }
        return result;
    }
}