[leetcode] 103.Binary Tree Zigzag Level Order Traversal
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题目:
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree [3,9,20,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7]]题解:
根据其特点我们用到栈的后进先出的特点,这道题我们维护两个栈,相邻两行分别存到两个栈中,进栈的顺序也不相同,一个栈是先进左子结点然后右子节点,另一个栈是先进右子节点然后左子结点,这样出栈的顺序就是我们想要的之字形了,
public List<List<Integer>> zigzagLevelOrder(TreeNode root){List<List<Integer>> lists=new ArrayList<List<Integer>>();Stack<TreeNode> stack1=new Stack<TreeNode>();Stack<TreeNode> stack2=new Stack<TreeNode>();if(root!=null)stack1.push(root);List<Integer> list=new ArrayList<Integer>();while(!stack1.isEmpty() || !stack2.isEmpty()){while(!stack1.isEmpty()){TreeNode node=stack1.pop();list.add(node.val);if(node.left!=null)stack2.push(node.left);if(node.right!=null)stack2.push(node.right);}if(!list.isEmpty()){lists.add(list);list=new ArrayList<Integer>();}while(!stack2.isEmpty()){TreeNode node=stack2.pop();list.add(node.val);if(node.right!=null)stack1.push(node.right);if(node.left!=null)stack1.push(node.left);}if(!list.isEmpty()){lists.add(list);list=new ArrayList<Integer>();}}return lists;}注意:最开始我做时一直报超时,原因是因为
我把while(!stack1.isEmpty() || !stack2.isEmpty()) 写成 while(stack1!=null || stack2!=null)
对象一经实例化(new),就会初始化该对象,并在内存为其分配地址,并不等于空
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