LeetCode | 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.
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */
//注意:这个题与昨天的题思想有些不同。对于平衡树,要求树的每个节点都满足平衡的性质,即每个节点的左右子节点高度差都<=1 //而这个题并不要求树的每个节点都满足对称性(否则就变相要求树必须是满树了),例如树{1 2 2 # 3 3 #}是对称树public class Solution { public boolean subtreeSymmetric(TreeNode left_root,TreeNode right_root){ //判断两棵子树是否是对称的函数 if(left_root==null && right_root==null) return true; if(left_root!=null && right_root==null) return false; if(left_root==null && right_root!=null) return false; //子树是对称的条件是子树的根完全一样 //此外,递归的比较左根的右子树与右根的左子树是否是对称的 if(left_root.val != right_root.val) return false; //以及左根的左子树与右根的右子树是否是对称的 return subtreeSymmetric(left_root.left,right_root.right) && subtreeSymmetric(left_root.right,right_root.left); } public boolean isSymmetric(TreeNode root) { if(root == null) return true; if(root.left==null && root.right==null) return true; if(root.left!=null && root.right==null) return false; if(root.left==null && root.right!=null) return false; if(root.left.val != root.right.val) return false; //一些基本的判断 return subtreeSymmetric(root.left,root.right); //逐级向下判断子树是否是对称的 }}
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