65. Symmetric Tree
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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
Note:
Bonus points if you could solve it both recursively and iteratively.
分析:题目要求判断给定的一个二叉树是不是对称二叉树。
/** * Step1:首先判断root是否为空结点或无孩子节点,若是则返回true; * Step2:判断root是否只有一个孩子,若是则返回false; * Step3:判断左右孩子结点的值是否相等,若不是则返回false; * Step4:调用isSameTree()方法,判断两个子树是否对称。 * */public boolean isSymmetric(TreeNode root) {if(root == null || root.left==null && root.right == null){return true;}else if(root.left==null || root.right == null){return false;}else{if(root.left.val != root.right.val){return false;}else{return isSymmetricTwoTree(root.left,root.right);}} }/** * 判断根节点的左右子树是否对称 */public boolean isSymmetricTwoTree(TreeNode p, TreeNode q) {/*两个都为null,返回true*/ if(p==null && q==null){ return true; }else if(p==null || q==null){/*只有一个为null。另一个不为null则返回false*/ return false; }else{ if(p.val == q.val){ return isSymmetricTwoTree(p.left,q.right) && isSymmetricTwoTree(p.right,q.left); }else{ return false; } } }
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