【LeetCode】 101. Symmetric Tree
2016-12-19 12:09
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Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree
But the following
Note:
Bonus points if you could solve it both recursively and iteratively.
/**
* 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 isSymmetricHelper(root.left, root.right);
}
private boolean isSymmetricHelper(TreeNode left, TreeNode right) {
if (left == null && right == null) {
return true;
}
if (left == null || right == null) {
return false;
}
if (left.val == right.val) {
return isSymmetricHelper(left.left, right.right) && isSymmetricHelper(le
4000
ft.right, right.left);
}
return false;
}
}
For example, this binary tree
[1,2,2,3,4,4,3]is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following
[1,2,2,null,3,null,3]is not:
1 / \ 2 2 \ \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
/**
* 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 isSymmetricHelper(root.left, root.right);
}
private boolean isSymmetricHelper(TreeNode left, TreeNode right) {
if (left == null && right == null) {
return true;
}
if (left == null || right == null) {
return false;
}
if (left.val == right.val) {
return isSymmetricHelper(left.left, right.right) && isSymmetricHelper(le
4000
ft.right, right.left);
}
return false;
}
}
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