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LeetCode-101. Symmetric Tree

2022-08-10 15:36:00 【51CTO】

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

But the following [1,2,2,null,3,null,3] is not:

Note:
Bonus points if you could solve it both recursively and iteratively.

题解：

       
       class 
       
       Solution {
       
       public:
       
       bool 
       
       bothAreSymmetric (
       
       TreeNode
       
       * 
       
       a, 
       
       TreeNode
       
       * 
       
       b) {
       
       if (
       
       a 
       
       == 
       
       NULL 
       
       && 
       
       b 
       
       == 
       
       NULL) {
       
       return 
       
       true;
       
    }
       
       else 
       
       if (
       
       a 
       
       != 
       
       NULL 
       
       && 
       
       b 
       
       != 
       
       NULL 
       
       && 
       
       a
       
       ->
       
       val 
       
       == 
       
       b
       
       ->
       
       val) {
       
       return 
       
       bothAreSymmetric(
       
       a
       
       ->
       
       left, 
       
       b
       
       ->
       
       right) 
       
       && 
       
       bothAreSymmetric(
       
       a
       
       ->
       
       right, 
       
       b
       
       ->
       
       left);
       
    }
       
       else {
       
       return 
       
       false;
       
    }
       
  }
       
       bool 
       
       isSymmetric(
       
       TreeNode
       
       * 
       
       root) {
       
       if (
       
       root 
       
       == 
       
       NULL) {
       
       return 
       
       true;
       
    }
       
       if (
       
       root
       
       ->
       
       left 
       
       == 
       
       NULL 
       
       || 
       
       root
       
       ->
       
       right 
       
       == 
       
       NULL) {
       
       return 
       
       root
       
       ->
       
       left 
       
       == 
       
       root
       
       ->
       
       right;
       
    }
       
       return 
       
       root
       
       ->
       
       left
       
       ->
       
       val 
       
       == 
       
       root
       
       ->
       
       right
       
       ->
       
       val 
       
       && 
       
       bothAreSymmetric(
       
       root
       
       ->
       
       left
       
       ->
       
       right, 
       
       root
       
       ->
       
       right
       
       ->
       
       left) 
       
       && 
       
       bothAreSymmetric(
       
       root
       
       ->
       
       left
       
       ->
       
       left, 
       
       root
       
       ->
       
       right
       
       ->
       
       right);
       
  }
       
};
      
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版权声明
本文为[51CTO]所创，转载请带上原文链接，感谢
https://blog.51cto.com/u_14150327/5564024

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