LeetCode(101) Symmetric Tree解题报告

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

递归解法:

public class Solution {       public boolean isSymmetric(TreeNode root) {       if(root == null)           return true;       return isSym(root.left,root.right);    }    public boolean isSym(TreeNode left,TreeNode right){        if(left == null && right == null)            return true;        if(left == null || right == null)            return false;        return left.val == right.val && isSym(left.left,right.right) && isSym(left.right,right.left);    }}

迭代解法:

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.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;    }

Note
递归的效率(1ms)要比迭代高，而迭代的话，左子树遍历先序，右子树遍历后序的效率(4ms)又要略低于左子树遍历后序，右子树遍历先序的效率(3ms)，不知道迭代的这个变化这个是不是跟后台数据有关。