题目来源

226. 翻转二叉树 - 力扣（LeetCode）

代码(dfs)

/**
 * 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) {
        if(root == null) return root;

        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);

        while(!stack.isEmpty()) {
            TreeNode node = stack.pop();

            if(node.right != null) stack.push(node.right);
            if(node.left != null) stack.push(node.left);

            swapNode(node);
        }
        return root;
    }
    //交换某结点左右子树
    private void swapNode(TreeNode node) {
        TreeNode tmp = node.left;
        node.left = node.right;
        node.right = tmp;
    }

}

代码分析

直接正常入栈就行，把栈顶结点直接交换左右子树就行。也就是，从根节点开始，交换左右子树，然后左节点，最后右结点。