Java 二叉树的遍历实现

前序遍历:先访问根节点,再访问左子节点,最后访问右子节点(根左右)

中序遍历:先访问左子节点,再访问根节点,最后访问右子节点(左根右)

后序遍历:先访问左子节点,再访问右子节点,最后访问根节点(左右根)

这三种遍历都有递归和循环两种实现方法,每种遍历的递归实现都比循环实现要简洁的多。

leetcode练习地址:

二叉树的前序遍历    

二叉树的中序遍历

二叉树的后序遍历

前序遍历的递归方法:

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

    private void printOne(TreeNode root,List<Integer> result){
        if(root != null){
            result.add(root.val);
            printOne(root.left, result);
            printOne(root.right, result);
        }
    }
}

前序遍历的循环方法(迭代方法,用栈放右子节点):

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

中序遍历和后序遍历的递归只需修改三个节点的执行顺序就可以了

中序改为

printOne(root.left, result);
result.add(root.val);
printOne(root.right, result);

后序改为

printOne(root.left, result);
printOne(root.right, result);
result.add(root.val);

中序迭代方法

public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<>();
    Deque<TreeNode> stack = new ArrayDeque<>();
    TreeNode p = root;
    while(!stack.isEmpty() || p != null) {
        if(p != null) {
            stack.push(p);
            p = p.left;
        } else {
            TreeNode node = stack.pop();
            result.add(node.val); 
            p = node.right;   
        }
    }
    return result;
}

后序迭代方法

public List<Integer> postorderTraversal(TreeNode root) {
    LinkedList<Integer> result = new LinkedList<>();
    Deque<TreeNode> stack = new ArrayDeque<>();
    TreeNode p = root;
    while(!stack.isEmpty() || p != null) {
        if(p != null) {
            stack.push(p);
            result.addFirst(p.val); 
            p = p.right;           
        } else {
            TreeNode node = stack.pop();
            p = node.left;     
        }
    }
    return result;
}


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转载自blog.csdn.net/whut2010hj/article/details/80770691