LEETCODE101 Translation C language version Python version __python

Symmetric tree

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

For example, this binary the 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. 101. Symmetric tree Given a binary tree, check whether it is its own mirror image (that is, axisymmetric) For example, this binary tree is symmetrical:

    1
   /\
  2   2
 /\/\
3  4  4 3

But this is not the following:

    1
   /\
  2   2
   \   \
   3    3

Train of thought: This problem solves with recursion very well, at the beginning will Saozi the right subtree of tree to go down separately, if the right subtree of the second tree of the first tree is Saozi exactly the same, and the right subtree of the first tree and the Zuozi of the second tree are recursively identical, then the two trees will satisfy symmetry. If the two Shang trees of the root knot satisfy symmetry, then the whole tree is symmetrical.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 * *
 
 bool Issymmetriclr (struct treenode* left, struct treenode* right) {
     if (left = null && right = NULL) return true;
     if (left = null && right!= null) return false;
     if (left!= null && right = NULL) return false;
     if (left->val!= right->val) return false;
     if (!ISSYMMETRICLR (Left->left, Right->right)) return false;
     if (!ISSYMMETRICLR (Left->right, Right->left)) return false;
     return true;
 }
 
BOOL Issymmetric (struct treenode* root) {
    if (root = NULL) return true;
    Return issymmetriclr (Root->left, root->right);
}

# Definition for a binary tree node.
# class TreeNode (object):
#     def __init__ (self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution (object):
    def issymmetric (self, Root): ""
        : Type root: TreeNode
        : Rtype:bool
        ""
        if not Root:return True
        to return self.issymmetriclr (Root.left, Root.right
        
    def issymmetriclr (self, left, right):
        if not (left or right): "Return True
        if" (left and Right==n One) or (Left==none and right): return
            False
        if Left.val!= right.val:return false
        if not self.issymmetric LR (Left.left, right.right): return false
        if not SELF.ISSYMMETRICLR (Left.right, Right.left): Return False
        Return True