leetcode101. Symmetric Tree

题目要求

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3
But the following [1,2,2,null,3,null,3] is not:
    1
   / \
  2   2
   \   \
   3    3
Note:
Bonus points if you could solve it both recursively and iteratively.

检查一棵树是否是左右对称的。

递归

在这里递归的一般情况是，输入进行比较的左子树和右子树的根节点，先判断该俩根节点是否等价，然后判断子节点是否等价。

    public boolean isSymmetric(TreeNode treeNode){
        if(treeNode==null) return true;
        return isSymmetric(treeNode.left, treeNode.right);
    }
    
    public boolean isSymmetric(TreeNode left, TreeNode right){
        if(left==null && right==null) return true;
        if(left!=null && right!=null && left.val==right.val){
            return isSymmetric(left.left, right.right)&&isSymmetric(left.right, right.left);
        }
        return false;
    }

栈

通过栈的形式同样可以实现比较。将需要进行比较的节点依次压入栈中。每次从栈中取出两个进行比较的节点比较。有点像level traversal的思路。

    public boolean isSymmetric2(TreeNode treeNode){
        if(treeNode==null) return true;
        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
        TreeNode left = treeNode.left, right = treeNode.right;
        stack.push(left);
        stack.push(right);
        while(!stack.isEmpty()){
            right = stack.pop();
            left = stack.pop();
            if(right==null && left==null)continue;
            else if(left==null || right==null)return false;
            else if(left.val==right.val){
                stack.push(left.left);
                stack.push(right.right);
                stack.push(left.right);
                stack.push(right.left);
            }else{
                return false;
            }
        }
        return true;
    }

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