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:
1 / 2 2 3 3 Note:
Bonus points if you could solve it both recursively and iteratively.
给定一棵树,推断它是否是对称的。
即树的左子树是否是其右子树的镜像。
使用递归进行求解。先推断左右子结点是否相等,不等就返回false。相等就将左子结点的左子树与右子结果的右子结点进行比較操作,同一时候将左子结点的左子树与右子结点的左子树进行比較,仅仅有两个同一时候为真是才返回true。否则返回false。
树结点类
public class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; }}算法实现类
public class Solution { public boolean isSymmetric(TreeNode root) { if (root == null) { return true; } else { return isSame(root.left, root.right); } } private boolean isSame(TreeNode left, TreeNode right) { if (left == null && right == null) { return true; } if (left != null && right == null || left == null && right != null){ return false; } else { return left.val == right.val && isSame(left.left, right.right) && isSame(left.right, right.left); } }} 评測结果点击图片,鼠标不释放。拖动一段位置。释放后在新的窗体中查看完整图片。
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