103. Binary Tree Zigzag Level Order Traversal
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]
]
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> listNode = new LinkedList<>();
Stack<TreeNode> stack1 = new Stack<>();
Stack<TreeNode> stack2 = new Stack<>();
if(root == null){
return listNode;
}
stack1.push(root);
while(!stack1.isEmpty()||!stack2.isEmpty()){
List<Integer> subList = new LinkedList<>();
if(!stack1.isEmpty()){
while(!stack1.isEmpty()){
TreeNode node = stack1.pop();
if(node!=null){
subList.add(node.val);
stack2.push(node.left);
stack2.push(node.right);
}
}
}else if(!stack2.isEmpty()){
while(!stack2.isEmpty()){
TreeNode node = stack2.pop();
if(node!=null){
subList.add(node.val);
stack1.push(node.right);
stack1.push(node.left);
}
}
}
if(!subList.isEmpty()){
listNode.add(subList);
}
}
return listNode;
}
}借助两个栈实现，注意出栈入栈的顺序