101. Symmetric Tree

题目101. Symmetric Tree

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.

1,利用递归

 public boolean isSymmetric(TreeNode root) {
        if(root == null){
            return true;
        }
        
        return dfs(root.left,root.right);
    }
    
    private boolean dfs(TreeNode left, TreeNode right){
        if(left == null && right == null){
            return true;
        }
        
        if(left == null || right == null){
            return false;
        }
        
        if(left.val == right.val){
            return dfs(left.left,right.right) && dfs(left.right,right.left);
        }else{
            return false;
        }
    }

2,利用"中序"遍历

思路:两次"中序"遍历, 左-根-右,   右-根-左
二者遍历的结果一样,则返回true,否则false

public boolean isSymmetric(TreeNode root) {
        if(root == null){
            return true;
        }
        StringBuffer forwardSb = new StringBuffer();
        StringBuffer reverseSb = new StringBuffer();
        Stack forwardStack = new Stack();
        Stack reverseStack = new Stack();
        TreeNode forwardNode = root;
        TreeNode reverseNode = root;
        do{
            while(forwardNode != null){
               forwardStack.push(forwardNode);
               forwardNode = forwardNode.left;
            }
            
            while(reverseNode != null){
               reverseStack.push(reverseNode);
               reverseNode = reverseNode.right;
            }
            
            if(forwardStack.size() != reverseStack.size()){
                return false;
            }
            
            if (!forwardStack.empty()){
                forwardNode = forwardStack.pop();
                reverseNode = reverseStack.pop();
                if(forwardNode.val != reverseNode.val){
                    return false;
                }
                
                forwardNode = forwardNode.right;
                reverseNode = reverseNode.left;
            }
        }while((!forwardStack.empty() || forwardNode != null) && (!reverseStack.empty() || reverseNode != null));
        
        return forwardStack.empty() && reverseStack.empty();
    }