LeetCode: Symmetric Tree 解题报告

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.

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     // solution 1:
     public boolean isSymmetric(TreeNode root) {
         if (root == null) {
             return true;
         }

         return isSymmetricTree(root.left, root.right);
     }

     /*
      * 判断两个树是否互相镜像
         (1) 根必须同时为空，或是同时不为空
      *
      * 如果根不为空:
         (1).根的值一样
         (2).r1的左树是r2的右树的镜像
         (3).r1的右树是r2的左树的镜像
      */
     public boolean isSymmetricTree1(TreeNode root1, TreeNode root2) {
         if (root1 == null && root2 == null) {
             return true;
         }

         if (root1 == null || root2 == null) {
             return false;
         }

         return root1.val == root2.val &&
              isSymmetricTree(root1.left, root2.right) &&
              isSymmetricTree(root1.right, root2.left);
     }

     // solution 2:
     public boolean isSymmetricTree(TreeNode root1, TreeNode root2) {
         if (root1 == null && root2 == null) {
             return true;
         }

         if (root1 == null || root2 == null) {
             return false;
         }

         Stack<TreeNode> s1 = new Stack<TreeNode>();
         Stack<TreeNode> s2 = new Stack<TreeNode>();

         // 一定记得初始化
         s1.push(root1);
         s2.push(root2);

         while (!s1.isEmpty() && !s2.isEmpty()) {
             TreeNode cur1 = s1.pop();
             TreeNode cur2 = s2.pop();

             if (cur1.val != cur2.val) {
                 return false;
             }

             if (cur1.left != null && cur2.right != null) {
                 s1.push(cur1.left);
                 s2.push(cur2.right);
             } else if (!(cur1.left == null && cur2.right == null)) {
                 return false;
             }

             if (cur1.right != null && cur2.left != null) {
                 s1.push(cur1.right);
                 s2.push(cur2.left);
             } else if (!(cur1.right == null && cur2.left == null)) {
                 return false;
             }
         }

         return true;
     }
 }

2015.1.20:

简化了递归的解法，root与root本身是一对对称树。

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public boolean isSymmetric(TreeNode root) {
         return isMirror(root, root);
     }

     public boolean isMirror(TreeNode r1, TreeNode r2) {
         if (r1 == null && r2 == null) {
             return true;
         } else if (r1 == null || r2 == null) {
             return false;
         }

         return r1.val == r2.val
              && isMirror(r1.left, r2.right)
              && isMirror(r1.right, r2.left);
     }
 }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsSymmetric.java