LeetCode-606. Construct String from Binary Tree (Java)

题目链接

此题是关于二叉树的，先回顾一下关于二叉树的相关知识

二叉树的先序遍历（根节点，左孩子，右孩子）：

public class PreTreeSolution {

    public static String preorderTraversal(TreeNode root) {
    	StringBuilder builder = new StringBuilder(); 
    	//用栈来存储节点
    	Stack stack = new Stack();
    	//根节点入栈
    	stack.push(root);
    	while(!stack.empty()){
    		//栈顶元素出栈
    		TreeNode n = stack.pop();
    		builder.append(n.val); 
    		//由于是 先序，所以让右节点先入栈
    		if(n.right != null){
    			stack.push(n.right);
    		}
    		//左节点再入栈，这样出栈的时候左节点就可以先出栈
    		if(n.left != null){
    			stack.push(n.left);
    		}
    	}
    	return builder.toString();
    }
    public static void main(String[] args) {
     	TreeNode rootNode = new TreeNode(1);
    	TreeNode node1 = new TreeNode(2);
    	TreeNode node2 = new TreeNode(3);
    	TreeNode node3 = new TreeNode(4);
    	TreeNode node4 = new TreeNode(5);
    	rootNode.left = node1;
    	rootNode.right = node2;
    	node1.left = node3;
    	node1.right = node4;
    	String resultString = preorderTraversal(rootNode);
    	System.out.println(resultString);
	}
}
class TreeNode{
	int val;
	TreeNode left;
	TreeNode right;
	public TreeNode(int x) {
		val = x;
	}
}

二叉树的中序遍历，（左孩子，根节点，右孩子）：

 public static String inorderTraversal(TreeNode root) {
    	StringBuilder builder = new StringBuilder(); 
    	if(root == null){
    		return null;
    	}
    	Stack stack = new Stack();
    	TreeNode p =root;
    	while(!stack.isEmpty() || p!=null){
    		if(p!=null){
    			stack.push(p);
    			p = p.left;
    		}
    		else{
    			TreeNode t = stack.pop();
    			builder.append(t.val);
    			p = t.right;
    		}
    	}
    	return builder.toString();
    }

二叉树的后序遍历（左孩子，右孩子，根节点）：

 public static String postOrderTraversal(TreeNode root) {
    	StringBuilder builder = new StringBuilder(); 
    	if(root == null){
    		return null;
    	}
    	Stack stack = new Stack();
    	//根节点入栈
    	stack.push(root);
    	while(!stack.isEmpty()){
    		//peek method : Looks at the object at the top of this stack without removing it from the stack.
    		TreeNode temp = stack.peek();
    		//如果节点左右孩子都为空，则出栈输出值
    		if(temp.left == null && temp.right == null){
    			TreeNode pop = stack.pop();
    			builder.append(pop.val);
    		}
    		else{
    			//如果右孩子不为空，则入栈
    			if(temp.right != null){
    				stack.push(temp.right);
    				//右孩子入栈后置为空。因为节点输出值时判断该节点的左右孩子是否为空，
    				//所以如果该节点的右孩子入栈后不置为空，则该节点出栈输出值时判断就会出现问题。
    				temp.right = null;
    			}
    			//如果左孩子不为空，则入栈，入栈后左孩子置为空，原因如上
    			if(temp.left != null){
    				stack.push(temp.left);
    				temp.left = null;
    			}
    		}
    	}
    	return builder.toString();
    }

更详细的内容请戳 参考链接

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则本道题对二叉树的遍历属于先序遍历。先看代码，本代码耗时24毫秒，beats 77.8%：

public static String tree2str(TreeNode t) {
    	StringBuilder builder = new StringBuilder(); 
    	if(t == null){
    		return null;
    	}
    	Stack