【leetcode】101. Symmetric Tree【java】递归和非递归两种方法

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 //方法1：递归
/*public class Solution {
    public boolean isSymmetric(TreeNode root) {
        return root == null || isSymmetricHelp(root.left, root.right);
    }
    private boolean isSymmetricHelp(TreeNode left, TreeNode right){
        if (left == null || right == null){
            return left == right;
        }
        if (left.val != right.val){
            return false;
        }
        return (isSymmetricHelp(left.left, right.right) && isSymmetricHelp(left.right, right.left));
    }
}*/
//方法2：非递归
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null){
            return true;
        }
        Stack stack = new Stack();
        stack.push(root.left);
        stack.push(root.right);
        while (!stack.empty()){
            TreeNode p1 = stack.pop();
            TreeNode p2 = stack.pop();
            if (p1 == null && p2 == null){
                continue;
            }
            if (p1 == null || p2 == null || p1.val != p2.val){
                return false;
            }
            stack.push(p1.left);
            stack.push(p2.right);
            stack.push(p1.right);
            stack.push(p2.left);
        }
        return true;
    }
}