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解题思路:前序遍历,然后将每个非空节点的左右子树交换(java要创建tmp变量交换),最后返回交换后的root;或者也可以创建一个新的二叉树节点然后赋值(比较麻烦)
class Solution {
public TreeNode invertTree(TreeNode root) {
if (root == null) return null;
treeReverse(root);
return root;
}
public void treeReverse(TreeNode root) {
if (root == null) return;
// 交换左右子节点
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
// 递归反转左右子树
treeReverse(root.left);
treeReverse(root.right);
}
}
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
return treeReverse(root);
}
public TreeNode treeReverse(TreeNode root){
if(root==null) return null;
TreeNode root_back = new TreeNode(root.val);
root_back.left = treeReverse(root.right);
root_back.right = treeReverse(root.left);
return root_back;
}
}