二叉树的锯齿形层次遍历 · Binary Tree Zigzag Level Order Traversal
[抄题]:
给出一棵二叉树,返回其节点值的锯齿形层次遍历(先从左往右,下一层再从右往左,层与层之间交替进行)
[思维问题]:
不知道反复切换要怎么做:用boolean normalOrder当作布尔型控制变量
[一句话思路]:
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
root一个点正常出入。第二层先左后右,出来就变成先右后左了。
[一刷]:
- 层遍历是放在原来的queue中,但是zigzag遍历的左右节点是放在下一层,另外一个栈next中。不要搞错
- 栈交换也是循环操作的,栈交换需要放在循环体里面
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[总结]:
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
用两个栈:可以实现逆序,来回交换
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
层遍历
public class Solution {
/*
* @param root: A Tree
* @return: A list of lists of integer include the zigzag level order traversal of its nodes' values.
*/
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
//define
Stack<TreeNode> curt = new Stack<TreeNode>();
Stack<TreeNode> next = new Stack<TreeNode>();
Stack<TreeNode> temp = new Stack<TreeNode>();
List<List<Integer>> result = new ArrayList<List<Integer>>();
boolean normalOrder = true; if (root == null) {
return result;
} //put into stack
curt.push(root);
while (!curt.isEmpty()) {
List<Integer> level = new LinkedList<Integer>();
int size = curt.size();
for (int i = 0; i < size; i++) {
TreeNode node = curt.pop();
level.add(node.val);
if (normalOrder == true) {
if (node.left != null) {
next.push(node.left);//
}
if (node.right != null) {
next.push(node.right);
}
}
else {
if (node.right != null) {
next.push(node.right);
}
if (node.left != null) {
next.push(node.left);
}
}
}
result.add(level);
//reverse stack
temp = curt;
curt = next;
next = temp;
normalOrder = !normalOrder;//
} //output result
return result;
}
}
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