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Merge pull request #110 from Joshua-Lu/patch-27

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更新 0101.对称二叉树 Java版本，递归与迭代
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Carl Sun
2021-05-14 10:57:38 +08:00
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parent af95c8ce09 cbf41ebf91
commit 9fc2f691c5
1 changed files with 79 additions and 64 deletions
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problems/0101.对称二叉树.md
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@ -254,93 +254,108 @@ public:
## 其他语言版本 ## 其他语言版本
Java： Java：
```java
public class N0101 {
/**
* 解法1：DFS，递归。
*/
public boolean isSymmetric2(TreeNode root) {
if (root == null) {
return false;
}
```Java
/**
* 递归法
*/
public boolean isSymmetric1(TreeNode root) {
return compare(root.left, root.right); return compare(root.left, root.right);
} }
private boolean compare(TreeNode left, TreeNode right) { private boolean compare(TreeNode left, TreeNode right) {
if (left == null && right == null) {
return true; if (left == null && right != null) {
return false;
} }
if (left != null && right == null) { if (left != null && right == null) {
return false; return false;
} }
if (left == null && right != null) {
if (left == null && right == null) {
return true;
}
if (left.val != right.val) {
return false; return false;
} }
// 比较外侧
if (left.val == right.val) { boolean compareOutside = compare(left.left, right.right);
return compare(left.left, right.right) && compare(left.right, right.left); // 比较内侧
} boolean compareInside = compare(left.right, right.left);
return compareOutside && compareInside;
return false;
} }
/** /**
* 解法2：DFS，迭代 * 迭代法
* 使用双端队列，相当于两个栈
*/ */
public boolean isSymmetric3(TreeNode root) { public boolean isSymmetric2(TreeNode root) {
if (root == null) { Deque<TreeNode> deque = new LinkedList<>();
return false; deque.offerFirst(root.left);
} deque.offerLast(root.right);
while (!deque.isEmpty()) {
if (!equal(root.left, root.right)) { TreeNode leftNode = deque.pollFirst();
return false; TreeNode rightNode = deque.pollLast();
} if (leftNode == null && rightNode == null) {
continue;
Deque<TreeNode> st = new LinkedList<>(); }
// if (leftNode == null && rightNode != null) {
st.push(root.right); // return false;
st.push(root.left); // }
// if (leftNode != null && rightNode == null) {
TreeNode curR = root.right; // return false;
TreeNode curL = root.left; // }
// if (leftNode.val != rightNode.val) {
while (!st.isEmpty()) { // return false;
curL = st.pop(); // }
curR = st.pop(); // 以上三个判断条件合并
if (leftNode == null || rightNode == null || leftNode.val != rightNode.val) {
// 前序，处理
if (!equal(curL, curR)) {
return false; return false;
} }
deque.offerFirst(leftNode.left);
if (curR != null && curL != null) { deque.offerFirst(leftNode.right);
st.push(curL.right); deque.offerLast(rightNode.right);
st.push(curR.left); deque.offerLast(rightNode.left);
st.push(curR.right);
st.push(curL.left);
}
} }
return true; return true;
} }
private boolean equal(TreeNode l, TreeNode r) { /**
if (l == null && r == null) { * 迭代法
return true; * 使用普通队列
*/
public boolean isSymmetric3(TreeNode root) {
Queue<TreeNode> deque = new LinkedList<>();
deque.offer(root.left);
deque.offer(root.right);
while (!deque.isEmpty()) {
TreeNode leftNode = deque.poll();
TreeNode rightNode = deque.poll();
if (leftNode == null && rightNode == null) {
continue;
}
// if (leftNode == null && rightNode != null) {
// return false;
// }
// if (leftNode != null && rightNode == null) {
// return false;
// }
// if (leftNode.val != rightNode.val) {
// return false;
// }
// 以上三个判断条件合并
if (leftNode == null || rightNode == null || leftNode.val != rightNode.val) {
return false;
}
// 这里顺序与使用Deque不同
deque.offer(leftNode.left);
deque.offer(rightNode.right);
deque.offer(leftNode.right);
deque.offer(rightNode.left);
} }
if (l != null && r == null) { return true;
return false;
}
if (l == null && r != null) {
return false;
}
if (l.val == r.val) {
return true;
}
return false;
} }
}
``` ```
Python： Python：