力扣每日一题-day1-翻转二叉树

最新推荐文章于 2024-08-18 14:30:00 发布

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最新推荐文章于 2024-08-18 14:30:00 发布

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分类专栏： Java 力扣

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在这里插入图片描述
答案：
Java

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

补充：前序遍历

public  void preOrder（TreeNode root）{
		if（root != null){
			System.out.print(root.value+”->”);
			preOrder(root.left);
			preOrder(root.right)
		}
}

非递归

  public void preOrderTraverse2(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        while (node != null || !stack.empty()) {
            if (node != null) {
                System.out.print(node.value + "->");
                stack.push(node);
                node = node.left;
            } else {
                TreeNode tem = stack.pop();
                node = tem.right;
            }
        }
    }

中序遍历

public  void preOrder（TreeNode root）{
		if（root != null){
			preOrder(root.left);
			System.out.print(root.value+”->”);
			preOrder(root.right)
		}
}

非递归

  public void preOrderTraverse2(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        while (node != null || !stack.empty()) {
            if (node != null) {
                stack.push(node);
                node = node.left;
            } else {
               TreeNode temp = stack.pop();
               System.out.print(temp.value+”->”);
               node = temp.right;
            }
        }
    }

后序遍历

public  void preOrder（TreeNode root）{
		if（root != null){
			preOrder(root.left);
			preOrder(root.right)
			System.out.print(root.value+”->”);
		}
}

非递归

 public void preOrderTraverse2(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
         TreeNode cur, pre = null;
          stack.push(root);
          
       while (!stack.empty()) {
            cur = stack.peek();
            if ((cur.left == null && cur.right == null) || (pre != null && (pre == cur.left || pre == cur.right))) {
                System.out.print(cur.val + "->");
                stack.pop();
                pre = cur;
            } else {
                if (cur.right != null)
                    stack.push(cur.right);
                if (cur.left != null)
                    stack.push(cur.left);
            }
        }
    }

执行演示

入栈 ：4 7 2 3 1                  pre = null, cur = 1
出栈：1  （4 7 2 3）              cur = 1, pre = 1
出栈：3  （4 7 2）                cur = 3，pre = 3
出栈：2  （4 7）                  cur = 2，pre = 2
入栈：9 ，6 （4 7 9 6）           cur = 7，pre = 2
出栈：6  （4 7 9）                cur = 6，pre = 6
出栈：9 （4 7）					 cur = 9 ，pre = 9
出栈：7，4
结束