107. Binary Tree Level Order Traversal II （二叉树由底向上层次遍历）

待我称王封你为后i 2022-07-15 01:18 61阅读 0赞

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,null,null,15,7]`,

3
       / \
      9  20
        /  \
       15   7

return its bottom-up level order traversal as:

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

/**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public List<List<Integer>> levelOrderBottom(TreeNode root) {
            List<List<Integer>> list = new LinkedList<List<Integer>>();
            if(root==null)
            	return list;
            List<Integer> intList = new LinkedList<Integer>();
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            TreeNode t = null;
            queue.offer(root);
            int curCount=0,curNum=1,nextCount=1;
            while(!queue.isEmpty()){
            	t=queue.poll();
            	intList.add(t.val);
            	if(t.left!=null){
            		queue.offer(t.left);
            		nextCount++;
            	}if(t.right!=null){
            		queue.offer(t.right);
            		nextCount++;
            	}
            	if(++curCount==curNum){
            		list.add(0,intList);
            		intList=new LinkedList<Integer>();
            		curNum=nextCount;
            	}
            }return list;
        }
    }