101. 对称二叉树

1,递归

条件是，两个叶子的值相当，左叶子的左叶子值等于右叶子的右叶子，左叶子的右叶子等于右叶子，空节点自然对称

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null){
            return true;
        }
        return   symmetric(root.left,root.right);
    }
    private boolean symmetric(TreeNode left, TreeNode right){
        if(left==null&&right==null){
            return true;
        }
        if(left==null||right==null){
            return false;
        }
        return ( (left.val==right.val)&&symmetric(left.left,right.right)&&symmetric(left.right,right.left) );
    }
}

另一种思路是用栈，这个只能自己多化几次图看看就明白了：

    public boolean isSymmetric(TreeNode root) {
        if(root==null){
            return true;
        }
        Stack<TreeNode> stack=new Stack<TreeNode>();
        TreeNode  pleft=root.left;
        TreeNode pright=root.right;
        stack.push(pleft);
        stack.push(pright);
        while(!stack.empty()){
            pleft=stack.pop();
            pright=stack.pop();
            if(pleft==null&&pright==null){
                continue;
            }
            if(pleft==null||pright==null){
                return false;
            }
            if(pleft.val!=pright.val){
                return false;
            }
            
            stack.push(pleft.left);
            stack.push(pright.right);
            stack.push(pleft.right);
            stack.push(pright.left);
        }
        return true;
    }