Symmetric Tree

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

SOL 1 & 2:

两个解法，递归和非递归.

1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     // solution 1:12     public boolean isSymmetric(TreeNode root) {13         if (root == null) {14             return true;15         }16         17         return isSymmetricTree(root.left, root.right);18     }19     20     /*21      * 判断两个树是否互相镜像 22         (1) 根必须同时为空，或是同时不为空23      * 24      * 如果根不为空: 25         (1).根的值一样 26         (2).r1的左树是r2的右树的镜像 27         (3).r1的右树是r2的左树的镜像28      */29     public boolean isSymmetricTree1(TreeNode root1, TreeNode root2) {30         if (root1 == null && root2 == null) {31             return true;32         }33         34         if (root1 == null || root2 == null) {35             return false;36         }37         38         return root1.val == root2.val && 39              isSymmetricTree(root1.left, root2.right) && 40              isSymmetricTree(root1.right, root2.left);41     }42     43     // solution 2:44     public boolean isSymmetricTree(TreeNode root1, TreeNode root2) {45         if (root1 == null && root2 == null) {46             return true;47         }48         49         if (root1 == null || root2 == null) {50             return false;51         }52         53         Stack
      
        s1 = new Stack
       
        ();54         Stack
        
          s2 = new Stack
         
          ();55         56         // 一定记得初始化57         s1.push(root1);58         s2.push(root2);59         60         while (!s1.isEmpty() && !s2.isEmpty()) {61             TreeNode cur1 = s1.pop();62             TreeNode cur2 = s2.pop();63             64             if (cur1.val != cur2.val) {65                 return false;66             }67             68             if (cur1.left != null && cur2.right != null) {69                 s1.push(cur1.left);70                 s2.push(cur2.right);71             } else if (!(cur1.left == null && cur2.right == null)) {72                 return false;73             }74             75             if (cur1.right != null && cur2.left != null) {76                 s1.push(cur1.right);77                 s2.push(cur2.left);78             } else if (!(cur1.right == null && cur2.left == null)) {79                 return false;80             }81         }82         83         return true;84     }85 }
         
        
       
      

 

2015.1.20:

简化了递归的解法，root与root本身是一对对称树。

1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     public boolean isSymmetric(TreeNode root) {12         return isMirror(root, root);13     }14     15     public boolean isMirror(TreeNode r1, TreeNode r2) {16         if (r1 == null && r2 == null) {17             return true;18         } else if (r1 == null || r2 == null) {19             return false;20         }21         22         return r1.val == r2.val23              && isMirror(r1.left, r2.right)24              && isMirror(r1.right, r2.left);25     }26 }

 

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