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IsSymmetric_LeetCode.java
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package Algorithms.tree;
import java.util.Stack;
public class IsSymmetric_LeetCode {
/*
* Given a binary tree, check whether it is a mirror of itself (ie,
* symmetric around its center).
*
* For example, this binary tree is symmetric:
*
* 1 / \ 2 2 / \ / \ 3 4 4 3
*
* But the following is not:
*
* 1 / \ 2 2 \ \ 3 3
*
* Note: Bonus points if you could solve it both recursively and
* iteratively.
*
* confused what "{1,#,2,3}" means? > read more on how binary tree is
* serialized on OJ.
*
* OJ's Binary Tree Serialization:
*
* The serialization of a binary tree follows a level order traversal, where
* '#' signifies a path terminator where no node exists below.
*
* Here's an example:
*
* 1 / \ 2 3 / 4 \ 5
*
* The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
*/
//
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
return isMirrorRec(root.left, root.right);
}
/*
* 判断两个树是否互相镜像
* (1) 根必须同时为空，或是同时不为空
*
* 如果根不为空:
* (1).根的值一样
* (2).r1的左树是r2的右树的镜像
* (3).r1的右树是r2的左树的镜像
*/
public boolean isMirrorRec(TreeNode r1, TreeNode r2) {
if (r1 == null && r2 == null) {
return true;
}
if (r1 == null || r2 == null) {
return false;
}
// should compare the value of the root. remember this.
return r1.val == r2.val
&& isMirrorRec(r1.left, r2.right)
&& isMirrorRec(r1.right, r2.left);
}
public boolean isMirror(TreeNode r1, TreeNode r2) {
if (r1 == null && r2 == null) {
return true;
}
if (r1 == null || r2 == null) {
return false;
}
// We can do preOrder traverse to judge if the trees are mirror.
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
s1.push(r1);
s2.push(r2);
while (!s1.isEmpty() && !s2.isEmpty()) {
// pop the current node out.
TreeNode cur1 = s1.pop();
TreeNode cur2 = s2.pop();
// Judge if the value of the node is equal.
if (cur1.val != cur2.val) {
return false;
}
// tree1的左节点，tree2的右节点，可以同时不为空，也可以同时为空，否则返回false.
if (cur1.left != null && cur2.right != null) {
s1.push(cur1.left);
s2.push(cur2.right);
} else if (!(cur1.left == null && cur2.right == null)) {
return false;
}
// tree1的右节点，tree2的左节点，可以同时不为空，也可以同时为空，否则返回false.
if (cur1.right != null && cur2.left != null) {
s1.push(cur1.right);
s2.push(cur2.left);
} else if (!(cur1.right == null && cur2.left == null)) {
return false;
}
}
return true;
}
}