思路:
一次前序遍历
public class Solution { public TreeNode invertTree(TreeNode root) { if(root == null) return root; TreeNode tmp = root.left; root.left = root.right; root.right = tmp; invertTree(root.left); invertTree(root.right); return root; } }
思路:
一次前序遍历
public class Solution { public TreeNode invertTree(TreeNode root) { if(root == null) return root; TreeNode tmp = root.left; root.left = root.right; root.right = tmp; invertTree(root.left); invertTree(root.right); return root; } }