(easy)LeetCode 226.Invert Binary Tree

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Invert a binary tree.

     4
   /     2     7
 / \   / 1   3 6   9

to

     4
   /     7     2
 / \   / 9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public  TreeNode invertTree(TreeNode root) {
		if (root == null || (root.left == null) && (root.right == null))
			return root;
		root=invertTreeLR(root,root.left, root.right);
		return root;
	}

	public  TreeNode invertTreeLR(TreeNode root,TreeNode left, TreeNode right) { //left,right位临时引用
		if (left == null && right != null) {
			left = right;
			right = null;
			left=invertTreeLR(left,left.left, left.right);
		} else if (left != null && right == null) {
			right = left;
			left = null;
			right=invertTreeLR(right,right.left, right.right);
		} else if (left == null && right == null){
			
		}
			
		else {
			TreeNode tmp = null;
			tmp = left;
			left = right;
			right = tmp;
			left=invertTreeLR(left,left.left, left.right);
			right=invertTreeLR(right,right.left, right.right);
			
			
		}
		root.left=left;
		root.right=right;
		return root;
	}
}

　　运行结果：

  

(easy)LeetCode 226.Invert Binary Tree

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原文地址：http://www.cnblogs.com/mlz-2019/p/4691548.html