leetcode -- 101. Symmetric Tree 【對稱樹,結構,內容】
題目
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.
題意
給定一個二叉樹,檢查它是否是對稱的。
分析及解答
- 【結構,內容】結構上要一致,同時內容上要相同。
方法1(遞迴):
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(非遞迴):
public boolean isSymmetric(TreeNode root) { Queue<TreeNode> q = new LinkedList<TreeNode>(); if(root == null) return true; q.add(root.left); q.add(root.right); while(q.size() > 1){ TreeNode left = q.poll(), right = q.poll(); if(left== null&& right == null) continue; if(left == null ^ right == null) return false; if(left.val != right.val) return false; q.add(left.left); q.add(right.right); q.add(left.right); q.add(right.left); } return true; }
方法2.2(較為複雜)
public boolean isSymmetric(TreeNode root) {
if(root==null) return true;
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode left, right;
if(root.left!=null){
if(root.right==null) return false;
stack.push(root.left);
stack.push(root.right);
}
else if(root.right!=null){
return false;
}
while(!stack.empty()){
if(stack.size()%2!=0) return false;
right = stack.pop();
left = stack.pop();
if(right.val!=left.val) return false;
if(left.left!=null){
if(right.right==null) return false;
stack.push(left.left);
stack.push(right.right);
}
else if(right.right!=null){
return false;
}
if(left.right!=null){
if(right.left==null) return false;
stack.push(left.right);
stack.push(right.left);
}
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