LeetCode | Symmetric Tree

题目

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.

题目要求递归（解法1）和非递归（解法2）都试试。

非递归其实就是用stack来辅助。

也有用中序遍历的方法，判断中序遍历结果是否对称。

解法1

public class SymmetricTree {
	public boolean isSymmetric(TreeNode root) {
		if (root == null) {
			return true;
		}
		return solve(root.left, root.right);
	}

	private boolean solve(TreeNode left, TreeNode right) {
		if (left == null && right == null) {
			return true;
		}
		if (left == null || right == null || left.val != right.val) {
			return false;
		}
		return solve(left.left, right.right) && solve(left.right, right.left);
	}
}

import java.util.Stack;

public class SymmetricTree {
	public boolean isSymmetric(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.isEmpty() || rightStack.isEmpty())) {
			TreeNode left = leftStack.pop();
			TreeNode right = rightStack.pop();
			if (left == null && right == null) {
				continue;
			}
			if (left == null || right == null || left.val != right.val) {
				return false;
			}
			leftStack.add(left.left);
			leftStack.add(left.right);
			rightStack.add(right.right);
			rightStack.add(right.left);
		}
		return true;
	}
}

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LeetCode | Symmetric Tree

原文：http://blog.csdn.net/perfect8886/article/details/21109137