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.

Iterative function:

 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         // Start typing your Java solution below
13         // DO NOT write main() function
14         LinkedList<TreeNode> left = new LinkedList<TreeNode>();
15         LinkedList<TreeNode> right = new LinkedList<TreeNode>();
16         if(root == null) return true;
17         else if(root.left == null && root.right == null) return true;
18         else if((root.left == null && root.right != null) 
　　　　　　　　　|| (root.left != null && root.right == null)) return false;
19         else{
20             left.add(root.left);
21             right.add(root.right);
22             while(!left.isEmpty() && !right.isEmpty()){
23                 TreeNode l = left.pop();
24                 TreeNode r = right.pop();
25                 if(l.val != r.val) return false;
26                 if(l.left != null && r.right != null){
27                     left.add(l.left);
28                     right.add(r.right);
29                 }else if((l.left != null && r.right == null) 
　　　　　　　　　　　　　　 || (l.left == null && r.right != null)){
30                     return false;
31                 }
32                 if(l.right != null && r.left != null){
33                     left.add(l.right);
34                     right.add(r.left);
35                 }else if((l.right != null && r.left == null) 
　　　　　　　　　　　　　　 || (l.right == null && r.left != null)){
36                     return false;
37                 }
38             }
39         }
40         return true;
41     }
42     
43     public void checkNode(){
44         
45     }
46 }

 

Recursive function:

 

 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     boolean rsl = true;
12     public boolean isSymmetric(TreeNode root) {
13         // Start typing your Java solution below
14         // DO NOT write main() function
15         rsl = true;
16         if(root == null) return rsl;
17         else{
18             checkNode(root.left,root.right);
19         }
20         return rsl;
21     }
22     
23     public void checkNode(TreeNode l, TreeNode r){
24         if(l == null && r == null ){
25             return;
26         }else if((l == null && r != null) || (l != null && r == null)){
27             rsl = false;
28             return;
29         }else{
30             checkNode(l.left, r.right);
31             checkNode(l.right,r.left);
32             if(l.val != r.val)rsl =false;
33             return;
34         }
35     }
36 }

 第二遍：

 1 public class Solution {
 2     public boolean isSymmetric(TreeNode root) {
 3         // IMPORTANT: Please reset any member data you declared, as
 4         // the same Solution instance will be reused for each test case.
 5         return root == null || compare(root.left, root.right);
 6     }
 7     public boolean compare(TreeNode t1, TreeNode t2){
 8         if(t1 == null && t2 == null) return true;
 9         else if((t1 == null || t2 == null) || (t1.val != t2.val))return false;
10         else return compare(t1.left, t2.right) && compare(t1.right, t2.left);
11     }
12 }

 

posted on 2013-09-25 03:05  Step-BY-Step  阅读(122)  评论(0)  编辑  收藏  举报

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