107. Binary Tree Level Order Traversal II

[Shawngbk]

代码实现与102题一样，只是把curList加入到list时作出了改变 把第index+1个以及第index+1个以后的元素都往后移一位，类似elementData[i+1]=elementData[i]; 然后再把obj放到index的位置。 当然，这个过程需要考虑index的合理性以及是否会溢出等问题。

/**

• 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> levelOrderBottom(TreeNode root) { List<List> list = new LinkedList<List>(); if(root == null) return list; Queue curLevel = new LinkedList(); curLevel.offer(root); while(!curLevel.isEmpty()) { int size = curLevel.size(); List curList = new LinkedList(); for(int i = 0; i < size; i++) { TreeNode curNode = curLevel.poll(); curList.add(curNode.val); if(curNode.left != null) { curLevel.offer(curNode.left); } if(curNode.right != null) { curLevel.offer(curNode.right); } } list.add(0, curList); } return list; } }