二叉树前序、中序、后序（迭代和递归方式）遍历

/**
 * 二叉树的节点定义
 * 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;
 *     }
 * }
 */

中序遍历

// 迭代

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

Kth Smallest Element in a BST

// 迭代
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Stack<TreeNode> stack = new Stack<>();
        while(!stack.isEmpty() || root != null){
            while(root != null){
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            k--; 
            if(k == 0) return root.val;
            root = root.right;
        }
        return -1;
    }
}

先序遍历

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

先序遍历应用

Validate Binary Search Tree

// 迭代方式
class Solution {
    public boolean isValidBST(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        long pre = Long.MIN_VALUE;   
        while(root != null || !stack.isEmpty()){
            while(root != null){
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            // System.out.print(root.val + " ");
            if(root.val <= pre) return false;
            pre = root.val;
            root = root.right;
        }
        return true;
    }
}
// 递归方式
class Solution {
    public boolean isValidBST(TreeNode root) {
        return isValidBSTHelper(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    
    public boolean isValidBSTHelper(TreeNode root, long minVal, long maxVal){
        if(root == null) return true;
        if(root.val <= minVal || root.val >= maxVal) return false;
        return isValidBSTHelper(root.left, minVal, root.val) && isValidBSTHelper(root.right, root.val, maxVal);
    }
}

后序遍历

// 递归方式
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        postorderHelper(root, list);
        return list;
    }
    public void postorderHelper(TreeNode root, List<Integer> list){
        if(root == null) return;
        postorderHelper(root.left, list);
        postorderHelper(root.right, list);
        list.add(root.val);
    }
    
}
// 迭代方式
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        List<Integer> list = new ArrayList<>();
        while(!stack.isEmpty() || root != null){
            if(root != null){
                list.add(0, root.val);
                stack.push(root);
                root = root.right;
            }else{
                root = stack.pop();
                root = root.left;
            }
        }
        return list;
    }
}

层次遍历

Binary Tree Level Order

// 递归方式
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ret = new ArrayList<>();
        if(root == null) return ret;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            List<Integer> list = new ArrayList<>();
            int size = queue.size();
            for(int i = 0; i < size; i++){
                TreeNode node = queue.poll();
                list.add(node.val);
                if(node.left != null) queue.offer(node.left);
                if(node.right != null) queue.offer(node.right);
            }
            ret.add(list);
        }
        return ret;
    }
}
// 迭代方式
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        levelOrderHelper(res, root, 0);
        return res;
    }
    public void levelOrderHelper(List<List<Integer>> res, TreeNode root, int height){
        if(root == null) return;
        if(res.size() <= height) res.add(new ArrayList<>());
        res.get(height).add(root.val);
        levelOrderHelper(res, root.left, height + 1);
        levelOrderHelper(res, root.right, height + 1);
    }
}

Binary Tree Zigzag Level Order Traversal

class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        if(root == null) return res;
        queue.offer(root);
        int rev = -1;
        while(!queue.isEmpty()){
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            for(int i = 0; i < size; i++){
                TreeNode node = queue.poll();
                if(rev == 1) list.add(0, node.val);
                else list.add(node.val);
                if(node.left != null) queue.offer(node.left);
                if(node.right != null) queue.offer(node.right);
            }
            res.add(list);
            rev = -rev;
        }
        return res;
    }
}