二叉树的深度遍历(Java)

前序遍历

1. 递归遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>(); //不需要写入递归里，只需要一个结果集即可
        preorder(root,res);
        return res;

    }
    public void preorder(TreeNode toot, List<Integer> res){
        if(toot == null){
            return; //千万不要忘记终止条件
        }
        res.add(toot.val);
        preorder(toot.left, res);
        preorder(toot.right, res);
    }
}

2. 迭代遍历

主要思想：利用栈，先放中间节点，取出并记录结果集。接着放右节点和左节点，并不断将栈口元素作为新的root

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

中序遍历

1. 递归遍历

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

2. 迭代遍历

主要思路：先看左侧（一直看到null），再看右侧

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        Stack<TreeNode> sk = new Stack<>();
        while(!sk.empty() || root != null){
            if(root != null){ //判断的是root而不是root.left
                sk.push(root); //栈中不断记录左侧和中间值
                root = root.left; //如果不为空先看左侧
            }
            else{
                root = sk.pop();
                res.add(root.val); //把最后一个有值的加入结果集
                root = root.right; //看右侧有没有值
            } 
        }      
        return res;
    }
}

后序遍历

1. 递归遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        postorder(root, res);
        return res;
    }
    public void postorder(TreeNode root,List<Integer> res){
        if(root == null){
            return; //退出方法
        }
        postorder(root.left, res); //只是起到了搜索的作用
        postorder(root.right, res);
        res.add(root.val);
    }

}

2. 迭代遍历

后序遍历参考前序遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) { 
        /**
        先按照前序来，再翻转
         */
        List<Integer> res = new ArrayList<>();
        if(root == null){
            return res;
        }
        Stack<TreeNode> sk = new Stack<>();
        sk.push(root.val);
        while(!sk.empty()){ 
            root = sk.pop();
            res.add(root.val);  
            if(root.left != null){
                sk.push(root.left);
            }
            if(root.right != null ){
                sk.push(root.right);
            }
        }
        Collections.reverse(res); //不要忘记翻转
        return res; 
    }
}