Question:
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
But the following is not:
My code:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null)
return true;
else
return isSymmetric(root.left, root.right);
}
private boolean isSymmetric(TreeNode left, TreeNode right) {
if (left == null && right == null)
return true;
else if (left != null && right == null)
return false;
else if (left == null && right != null)
return false;
else {
return (left.val == right.val) && isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
}
}
}
My test result:
**
總結(jié)：
這次作業(yè)也不是很難吧。
所以一個(gè)小時(shí)做完了兩道 easy級(jí)別的題目曼验。
真沒(méi)啥總結(jié)的纤泵，就是遞歸吧悔常。寫(xiě)遞歸的能力被之前上算法課培訓(xùn)出來(lái)了，比較簡(jiǎn)單的都還是能寫(xiě)寫(xiě)的月帝。哪替。。
加油豪娜。
**
Good luck, Richardo!
My code:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null)
return true;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root);
while (!q.isEmpty()) {
int size = q.size();
ArrayList<TreeNode> l = new ArrayList<TreeNode>(size);
while (!q.isEmpty())
l.add(q.poll());
for (int i = 0; i < size / 2; i++) {
TreeNode left = l.get(i);
TreeNode right = l.get(size - i - 1);
if (left.val != right.val)
return false;
if (left.left == null && right.right != null)
return false;
if (left.left != null && right.right == null)
return false;
if (left.right == null && right.left != null)
return false;
if (left.right != null && right.left == null)
return false;
}
for (int i = 0; i < size; i++) {
TreeNode temp = l.get(i);
if (temp.left != null)
q.offer(temp.left);
if (temp.right != null)
q.offer(temp.right);
}
if (q.size() % 2 == 1)
return false;
}
return true;
}
}
用層序遍歷實(shí)現(xiàn)了非遞歸解法。但是時(shí)間好長(zhǎng)哟楷。
用一個(gè)隊(duì)列侵歇，存放已經(jīng)被檢查過(guò)的結(jié)點(diǎn)，按從左往右的順序吓蘑。
然后每次進(jìn)循環(huán)惕虑，把他們輸入進(jìn)一個(gè)list，然后從兩端往中間磨镶，一個(gè)結(jié)點(diǎn)一個(gè)結(jié)點(diǎn)得掃描溃蔫。判斷他們的下一層在結(jié)構(gòu)上是否滿(mǎn)足對(duì)稱(chēng)。還要檢查他們自己的value是否相等琳猫。
如果都滿(mǎn)足條件伟叛。按序把它們的下一層輸入進(jìn)隊(duì)列。
然后就差不多了脐嫂。
遞歸做法：
My code:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null)
return true;
else if (root.left == null && root.right == null)
return true;
else if (root.left == null || root.right == null)
return false;
return isSymmetric(root.left, root.right);
}
private boolean isSymmetric(TreeNode left, TreeNode right) {
if (left.val != right.val)
return false;
boolean b1 = false;
boolean b2 = false;
if (left.left == null && right.right == null)
b1 = true;
else if (left.left == null || right.right == null)
return false;
else
b1 = isSymmetric(left.left, right.right);
if (!b1)
return false;
if (left.right == null && right.left == null)
b2 = true;
else if (left.right == null || right.left == null)
return false;
else
b2 = isSymmetric(left.right, right.left);
return b2;
}
}
意思差不多统刮。
昨晚打電話(huà)到兩點(diǎn)。又早起上課账千，一直沒(méi)停侥蒙。所以晚上的時(shí)候特別累。還得跟著去上課匀奏。所以人比較暴躁鞭衩，心態(tài)不好。
晚上回去好好休息娃善。
又拿到了一個(gè)OA论衍。好好珍惜。
最近開(kāi)始練習(xí) Tree聚磺。
Anyway, Good luck, Richardo!
My code:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
return helper(root.left, root.right);
}
private boolean helper(TreeNode left, TreeNode right) {
if (left == null && right == null) {
return true;
}
else if (left == null || right == null) {
return false;
}
else if (left.val != right.val) {
return false;
}
else {
return helper(left.left, right.right) && helper(left.right, right.left);
}
}
}
recursion 不是很難坯台。
看了上面的 iteration解法，感覺(jué)有點(diǎn)煩瘫寝。
然后看了下答案：
My code:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.add(root);
q.add(root);
while (!q.isEmpty()) {
TreeNode n1 = q.poll();
TreeNode n2 = q.poll();
if (n1 == null && n2 == null) {
continue;
}
else if (n1 == null || n2 == null) {
return false;
}
else if (n1.val != n2.val) {
return false;
}
else {
q.offer(n1.left);
q.offer(n2.right);
q.offer(n1.right);
q.offer(n2.left);
}
}
return true;
}
}
reference:
https://leetcode.com/articles/symmetric-tree/
這種層序遍歷和之前的不太一樣蜒蕾。他是每次壓入兩個(gè)，彈出兩個(gè)矢沿。
Interesting
當(dāng)分析 recursion 復(fù)雜度時(shí)滥搭，
分析空間的時(shí)候，記住考慮到 recursive call 需要在棧中占的體積捣鲸。
比如這道題目瑟匆，在最?lèi)毫拥那闆r下，如果樹(shù)是一條線(xiàn)栽惶，那么它所占的體積就是 O(n)
Anyway, Good luck, Richardo! -- 08/28/2016
