144. Binary Tree Preorder Traversal

题目：

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

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连接： http://leetcode.com/problems/binary-tree-preorder-traversal/

题解：

二叉树先序遍历， root -> left -> right。使用一个栈来维护已经访问过的节点。当root不为空时，当前节点入栈，输出节点值，继续向左子树遍历。当root为空，从栈中弹出节点，向右子树进行遍历。

Iterative:

Time Complexity - O(n)，  Space Complexity - O(n)。

public class Solution {
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(root == null)
            return result;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        
        while(root != null || !stack.isEmpty()){
            if(root != null){
                stack.push(root);
                result.add(root.val);
                root = root.left;
            } else {
                root = stack.pop();
                root = root.right;
            }
        }
        
        return result;
    }
}

Recursive:

Time Complexity - O(n)， Space Complexity - O(n)。

public class Solution {
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(root == null)
            return result;
        helper(result, root);
        return result;
    }
    
    private void helper(ArrayList<Integer> result, TreeNode root){
        if(root == null)
            return;
        result.add(root.val);
        helper(result, root.left);
        helper(result, root.right);
    }
}

Update:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null)
            return res;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        
        while(!stack.isEmpty()) {
            root = stack.pop();
            if(root != null) {
                res.add(root.val);
                if(root.right != null)
                    stack.push(root.right);
                if(root.left != null)
                    stack.push(root.left);
            }
        }
        
        return res;
    }
}

Morris-Travel: 待定。

二刷:

Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Stack<TreeNode> stack = new Stack<>();
        while (root != null || !stack.isEmpty()) {
            if (root != null) {
                res.add(root.val);
                stack.push(root.right);
                root = root.left;
            } else {
                root = stack.pop();
            }
        }
        return res;
    }
}

三刷:

Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        
        while (!stack.isEmpty() || root != null) {
            if (root != null) {
                res.add(root.val);
                stack.push(root.right);
                root = root.left;
            } else {
                root = stack.pop();
            }
        }
        return res;
    }
}

Recursive:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        preorderTraversal(res, root);
        return res;
    }
    
    private void preorderTraversal(List<Integer> res, TreeNode root) {
        if (root == null) return;
        res.add(root.val);
        preorderTraversal(res, root.left);
        preorderTraversal(res, root.right);
    }
}

三刷:

Java:

Iterative with a stack:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        TreeNode node = root;
        Stack<TreeNode> stack = new Stack<>();
        while (node != null || !stack.isEmpty()) {
            if (node != null) {
                res.add(node.val);
                stack.push(node);
                node = node.left;
            } else {
                node = stack.pop();
                node = node.right;
            }
        }
        return res;
    }
}

Recursive:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        preorderTraversal(res, root);
        return res;
    }
    
    private void preorderTraversal(List<Integer> res, TreeNode root) {
        if (root == null) return;
        res.add(root.val);
        preorderTraversal(res, root.left);
        preorderTraversal(res, root.right);
    }
}

Morris traversal:

使用了类似in-order Morris Traversal类似的代码。在可以改变树结构的情况下，使用空的右子树来连接到父节点，最后再还原树结构。下面详述一下步骤：

  1. 在当前node非空的情况下对树进行遍历。仍然是先判断左子树是否为空，假如为空，则我们将当前节点node.val加入到结果集里，继续向右遍历树
  2. 否则左子树非空，这时我们向左子树里查找当前节点node的predecessor，简称prev，即左子树中最右端的元素。依然是两种情况：
    1. 当prev.right为空时，我们第一次访问此节点，这时把prev.right设置为当前节点node，然后把当前节点node.val加入到结果集里，向左遍历树
    2. 否则prev.right = 为当前node，说明我们是第二次访问这个节点，这时要做一个树的还原操作，也就是断开prev和node的连接，即prev.right = null，然后向右遍历树。这时候因为我们的当前节点node已经被处理过了，所以不要再次处理node.val 
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        TreeNode node = root, prev = null;
        while (node != null) {
            if (node.left == null) {
                res.add(node.val);
                node = node.right;
            } else {
                prev = node.left;
                while (prev.right != null && prev.right != node) {
                    prev = prev.right;
                }
                if (prev.right == null) {
                    prev.right = node;
                    res.add(node.val);
                    node = node.left;
                } else {
                    prev.right = null;
                    node = node.right;
                }
            }
        }
        return res;
    }
}

Update:

简化了一下，反而比较古怪，不易理解。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        TreeNode node = root, prev = null;
        while (node != null) {
            prev = node.left;
            while (prev != null && prev.right != null && prev.right != node) {
                prev = prev.right;
            }
            if (prev != null && prev.right == null) {
                prev.right = node;
                res.add(node.val);
                node = node.left;
            } else {
                if (prev != null) prev.right = null;
                else res.add(node.val);
                node = node.right;
            }    
        }
        return res;
    }
}

Reference：

https://leetcode.com/discuss/9734/accepted-code-explaination-with-algo

https://leetcode.com/discuss/5331/whats-the-simplest-and-cleanest-answer

https://leetcode.com/discuss/23326/very-simple-iterative-python-solution

https://leetcode.com/discuss/19798/accepted-iterative-solution-in-java-using-stack

https://leetcode.com/discuss/32904/3-different-solutions

https://leetcode.com/discuss/49926/easy-read-java-solutions-for-both-iterative-recursive-300ms

https://leetcode.com/discuss/46894/java-solution-both-recursion-and-iteration 

94. Binary Tree Inorder Traversal

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原文地址：https://www.cnblogs.com/yrbbest/p/4483758.html