leetcode151: symmetric tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

给定一颗二叉树，判断该二叉树是否是对称的。

For example, this binary tree is symmetric:

如下面的二叉树是对称的。

1

/ \

2   2

/     \

3       3

Note:

Bonus points if you could solve it both recursively and iteratively.

提示：是否可以同时用递归和迭代的方法解决？

方法一：递归

public class SymmetricTree {
static class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
public boolean isSymmetric(TreeNode root) {
if(root == null) return true;
return isSymmetric(root.left, root.right);
}
public boolean isSymmetric(TreeNode leftNode, TreeNode rightNode){
if(leftNode == null && rightNode == null ) return true;
if(leftNode == null || rightNode == null) return false;
if(leftNode.val != rightNode.val) return  false;
boolean isLeft = isSymmetric(leftNode.left, rightNode.right);
boolean isRight = isSymmetric(leftNode.right, rightNode.left);
return isLeft && isRight;
}

public static void main(String args[]){
TreeNode root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(2);
root.left.left = new TreeNode(3);
root.left.right = new TreeNode(4);
root.right.left = new TreeNode(4);
root.right.right = new TreeNode(3);
System.out.println(new SymmetricTree().isSymmetric(root));
System.out.println(new SymmetricTree().isSymmetricIter(root));
}
}

方法二：迭代

public boolean isSymmetricIter(TreeNode root) {
if(root == null) return true;
Stack<TreeNode> leftStack = new Stack<TreeNode>();
Stack<TreeNode> rightStack = new Stack<TreeNode>();
leftStack.push(root.left);
rightStack.push(root.right);

while (leftStack.size() > 0 && rightStack.size() > 0){
TreeNode left = leftStack.pop();
TreeNode right = rightStack.pop();
if(left == null && right == null) continue;
if(left == null || right == null) return false;
if(left.val == right.val) {
leftStack.push(left.right);
leftStack.push(left.left);
rightStack.push(right.left);
rightStack.push(right.right);
}else{
return false;
}
}
return true;
}