題目連結

此題是關於二叉樹的，先回顧一下關於二叉樹的相關知識

二叉樹的先序遍歷（根節點，左孩子，右孩子）：

public class PreTreeSolution {

    public static String preorderTraversal(TreeNode root) {
    	StringBuilder builder = new StringBuilder(); 
    	//用棧來儲存節點
    	Stack<TreeNode> stack = new Stack<TreeNode>();
    	//根節點入棧
    	stack.push(root);
 
    	while(!stack.empty()){
    		//棧頂元素出棧
    		TreeNode n = stack.pop();
    		builder.append(n.val); 
    		//由於是 先序，所以讓右節點先入棧
    		if(n.right != null){
    			stack.push(n.right);
    		}
    		//左節點再入棧，這樣出棧的時候左節點就可以先出棧
    		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;
	}
}
 

 public static String inorderTraversal(TreeNode root) {
    	StringBuilder builder = new StringBuilder(); 
    	if(root == null){
    		return null;
    	}
    	Stack<TreeNode> stack = new Stack<TreeNode>();
    	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();
    }

 public static String postOrderTraversal(TreeNode root) {
    	StringBuilder builder = new StringBuilder(); 
    	if(root == null){
    		return null;
    	}
    	Stack<TreeNode> stack = new Stack<TreeNode>();
    	//根節點入棧
    	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();
    		//如果節點左右孩子都為空，則出棧輸出值
    		if(temp.left == null && temp.right == null){
    			TreeNode pop = stack.pop();
    			builder.append(pop.val);
    		}
    		else{
    			//如果右孩子不為空，則入棧
    			if(temp.right != null){
    				stack.push(temp.right);
    				//右孩子入棧後置為空。因為節點輸出值時判斷該節點的左右孩子是否為空，
    				//所以如果該節點的右孩子入棧後不置為空，則該節點出棧輸出值時判斷就會出現問題。
    				temp.right = null;
    			}
    			//如果左孩子不為空，則入棧，入棧後左孩子置為空，原因如上
    			if(temp.left != null){
    				stack.push(temp.left);
    				temp.left = null;
    			}
    		}
    	}
    	return builder.toString();
    }

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則本道題對二叉樹的遍歷屬於先序遍歷。先看程式碼，本程式碼耗時24毫秒，beats 77.8%：

public static String tree2str(TreeNode t) {
    	StringBuilder builder = new StringBuilder(); 
    	if(t == null){
    		return null;
    	}
    	Stack<Object> stack = new Stack<Object>();
    	//根節點入棧
    	stack.push(t);
    	while(!stack.empty()){
    		//棧中包含TreeNode型別的二叉樹節點，也包含String型別的括號
    		Object temp = stack.pop();
    		//判斷是否為TreeNode型別
    		if(temp instanceof TreeNode){
    			TreeNode node = (TreeNode) temp;
    			builder.append(node.val);
    			//如果該節點左右節點都為空，則不執行任何程式碼，否則，執行以下程式碼
    			if(node.left != null || node.right != null){
    				if(node.right != null){
            			stack.push(")");
            			stack.push(node.right);
            			stack.push("(");
            		}
            		if(node.left != null){
            			stack.push(")");
            			stack.push(node.left);
            			stack.push("(");
            		}else{
            			stack.push(")");
            			stack.push("(");
            		}

    			}
    		}
    		//判斷是否為String型別
    		if(temp instanceof String){
    			//String型別直接拼接到builder
    			builder.append((String) temp);
    		}
       	}
    	return builder.toString();
    }