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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20, 9],
[15, 7]
]
分析:
层序遍历的变种,每层反向,可用栈代替队列来实现反向。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(root==null) return result;
Stack<TreeNode> st1 = new Stack<TreeNode>();
Stack<TreeNode> st2 = new Stack<TreeNode>();
st1.push(root);
while(st1.size()>0 || st2.size()>0 ){
List<Integer> level = new ArrayList<Integer>();
if(st1.size()>0){
while(st1.size()>0){
TreeNode temp = st1.pop();
level.add(temp.val);
if(temp.left != null) st2.push(temp.left);
if(temp.right != null) st2.push(temp.right);
}
result.add(level);
continue;
}
if(st2.size()>0){
while(st2.size()>0){
TreeNode temp = st2.pop();
level.add(temp.val);
if(temp.right!= null) st1.push(temp.right);
if(temp.left != null) st1.push(temp.left);
}
result.add(level);
}
}
return result;
}
}