【二叉树】剑指offer——面试题19：二叉树的镜像

力扣

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flipTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if not root:
            return None
        sub_left = self.flipTree(root.right)
        sub_right = self.flipTree(root.left)
        root.left = sub_left
        root.right = sub_right
        return root

Java

class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if (root == null || (root.left == null && root.right == null)) {
            return root;
        }
        if (root.left == null) {
            root.left = mirrorTree(root.right);
            root.right = null;
        } else if (root.right == null) {
            root.right = mirrorTree(root.left);
            root.left = null;
        } else {
            TreeNode originL = mirrorTree(root.left);
            TreeNode originR = mirrorTree(root.right);
            root.left = originR;
            root.right = originL;
        }
        
        return root;
    }
}

Solution1：
递归解法，牢记！
根结点为空的情况容易漏，注意！

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {//递归
public:
    void Mirror(TreeNode *pRoot) {
        if(pRoot == 0)//若整棵树的根结点为空则return
            return;
        if(pRoot->left == NULL && pRoot->right == NULL)//到达叶子结点，return
            return;
        
        struct TreeNode *temp;
        if(pRoot->left != NULL || pRoot->right != NULL){//存在子树，则交换之
            temp = pRoot->left;
            pRoot->left = pRoot->right;
            pRoot->right = temp;
        }
        if(pRoot->left)
            Mirror(pRoot->left);
        if(pRoot->right)
            Mirror(pRoot->right);
    }
};