Invert a binary tree.
Example 1:
Input: {1,3,#}
Output: {1,#,3}
Explanation:
1 1
/ => 3 3
Example 2:
Input: {1,2,3,#,#,4}
Output: {1,3,2,#,4}
Explanation:
1 1
/ \ / 2 3 => 3 2
/ 4 4
Do it in recursion is acceptable, can you do it without recursion?
/** * 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) { if (root == null) { return; } TreeNode temp = root.right; root.right = root.left; root.left = temp; invertBinaryTree(root.left); invertBinaryTree(root.right); } }
Lintcode175-Revert Binary Tree-Easy
原文：https://www.cnblogs.com/Jessiezyr/p/10661845.html