描述

226. 翻转二叉树 - 力扣（Leetcode）

分析

DFS的前序是最方便的

实现

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import java.util.Queue;
import java.util.LinkedList;

//226.翻转二叉树
public class InvertBinaryTree226 {
	public static void main(String[] args) {
			TreeNode root = buildTree(new int[]{1,2,3,4,5,6,7});
			root = new Solution().invertTree(root);
			levelOrderTraversal(root);

	}
	/**
	 *二叉树的构建
	 * int[] -> 二叉树
	 */
	 public static TreeNode buildTree(int[] arr) {
        return buildTreeHelper(arr, 0);
    }

    public static TreeNode buildTreeHelper(int[] arr, int index) {
        if (index >= arr.length) {
            return null;
        }

        TreeNode node = new TreeNode(arr[index]);

        node.left = buildTreeHelper(arr, 2 * index + 1);
        node.right = buildTreeHelper(arr, 2 * index + 2);

        return node;
    }

    /**
	 *	层序遍历
     */
    public static void levelOrderTraversal(TreeNode root) {
        if (root == null) {
            return;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            System.out.print(node.val + " ");

            if (node.left != null) {
                queue.add(node.left);
            }

            if (node.right != null) {
                queue.add(node.right);
            }
        }
    }
}

class Solution {
    public TreeNode invertTree(TreeNode root) {
    	if (root == null) {
    		return root;
    	}
    	invert(root);
    	invert(root.left);
    	invert(root.right);

    	return root;
    }
    void invert(TreeNode root){
    	TreeNode temp = root.left;
    	root.left = root.right;
    	root.right = temp;
    }
}

class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;

	public TreeNode(){};
	public TreeNode(int val){this.val = val;}
	public TreeNode(int val, TreeNode left, TreeNode right){
		this.val = val;
		this.left = left;
		this.right = right;
	}
}

总结