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

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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<TreeNode> stack = new Stack<TreeNode>();        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;    }}

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