描述
給出一棵二叉樹(shù),返回其節(jié)點(diǎn)值的鋸齒形層次遍歷(先從左往右,下一層再?gòu)挠彝螅瑢优c層之間交替進(jìn)行)
樣例
給出一棵二叉樹(shù) {3,9,20,#,#,15,7}
3
/ \
9 20
/ \
15 7
返回其鋸齒形的層次遍歷為:
[
[3],
[20,9],
[15,7]
]
代碼
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
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) {
List<List<Integer>> result = new ArrayList<>();
if (root == null) {
return result;
}
Stack<TreeNode> currLevel = new Stack<TreeNode>();
Stack<TreeNode> nextLevel = new Stack<TreeNode>();
// 不需要新建棧
Stack<TreeNode> tmp;
currLevel.push(root);
// normal代表當(dāng)前層順序是不是從左往右
boolean normalOrder = true;
// 因?yàn)閏urLevel和nextLevel不斷來(lái)回交換,第一個(gè)while用于判斷樹(shù)是否遍歷完最后一層
while (!currLevel.isEmpty()) {
// 每一層都要新建一個(gè)動(dòng)態(tài)數(shù)組存儲(chǔ)當(dāng)前層的值
ArrayList<Integer> currLevelResult = new ArrayList<Integer>();
// 第二個(gè)while用于判斷當(dāng)前層結(jié)點(diǎn)是否遍歷完成
while (!currLevel.isEmpty()) {
TreeNode node = currLevel.pop();
currLevelResult.add(node.val);
// 棧的彈出是反的,當(dāng)前層正常順序,下一層先壓左后壓右,先彈右后彈左
if (normalOrder) {
if (node.left != null) {
nextLevel.push(node.left);
}
if (node.right != null) {
nextLevel.push(node.right);
}
} else {
if (node.right != null) {
nextLevel.push(node.right);
}
if (node.left != null) {
nextLevel.push(node.left);
}
}
}
result.add(currLevelResult);
// 交換兩個(gè)棧的引用指向,當(dāng)前棧要pop成空棧,子結(jié)點(diǎn)加入到nextLevel
tmp = currLevel;
currLevel = nextLevel;
nextLevel = tmp;
normalOrder = !normalOrder;
}
return result;
}
}