lintcode ：Invert Binary Tree 翻转二叉树

题目：

翻转二叉树

翻转一棵二叉树

样例

  1         1
 /        / 
2   3  => 3   2
   /       
  4         4

挑战

递归固然可行，能否写个非递归的？

解题:

递归比较简单，非递归待补充

Java程序：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    public void invertBinaryTree(TreeNode root) {
        // write your code here
        invertTree(root);
        
    }
    public TreeNode invertTree(TreeNode root){
        if( root ==null){
            return root;
        }
        TreeNode rleft= root.left;
        root.left = invertTree(root.right);
        root.right = invertTree(rleft);
        return root;
    }
}

总耗时: 994 ms

Python程序：

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""
class Solution:
    # @param root: a TreeNode, the root of the binary tree
    # @return: nothing
    def invertBinaryTree(self, root):
        # write your code here
        self.invertTree(root)
        
    def invertTree(self,root):
        if root == None:
            return root
        rlft = root.left
        root.left = self.invertTree(root.right)
        root.right = self.invertTree(rlft)
        return root 

总耗时: 118 ms

参考剑指OfferP127改进了下程序，更容易理解了

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    public void invertBinaryTree(TreeNode root) {
        // write your code here
        invertTree(root);
    }
    public void invertTree(TreeNode root){
        if( root ==null){
            return;
        }
        if(root.left == null && root.right ==null){
            return;
        }
        TreeNode tmp= root.left;
        root.left = root.right;
        root.right = tmp;
        if(root.left!=null){
            invertTree(root.left);
        }
        if(root.right!=null){
            invertTree(root.right);
        }
    }
}

 

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""
class Solution:
    # @param root: a TreeNode, the root of the binary tree
    # @return: nothing
    def invertBinaryTree(self, root):
        # write your code here
        if root == None:
            return
        if root.left==None and root.right ==None:
            return
        tmp = root.left
        root.left = root.right
        root.right = tmp
        if root.left!=None:
            self.invertBinaryTree(root.left)
        if root.right!=None:
            self.invertBinaryTree(root.right)