[Index]Binary Tree Iterative Traversal

最新推荐文章于 2022-07-02 16:50:24 发布

原创最新推荐文章于 2022-07-02 16:50:24 发布 · 425 阅读

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本文详细介绍了二叉树的前序、中序和后序遍历算法，并提供了验证预序遍历序列是否为二叉搜索树的方法。

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    public List preorderTraversal(TreeNode root) {
        Stack stack = new Stack();
        List res = new ArrayList();
        TreeNode node = root;
        
        while(node != null || !stack.isEmpty()){
            if(node != null){
                stack.push(node);
                res.add(node.val);
                node = node.left;
            }else{
                TreeNode top = stack.pop();
                if(top.right != null){
                    node = top.right;
                }
            }
        }
        return res;
    }
    
    public List inorderTraversal(TreeNode root) {
        Stack stack = new Stack();
        List res = new ArrayList();
        
        while(root != null || !stack.isEmpty()){
            if(root != null){
                stack.push(root);
                root = root.left;
            }else{
                TreeNode top = stack.pop();
                res.add(top.val);
                root = top.right;
            }
        }
        return res;
    }
    
    public List postorderTraversal(TreeNode root) {
        Stack stack = new Stack();
        List res = new ArrayList();
        TreeNode prev = null;
        while(root != null || !stack.isEmpty()){
            if(root != null){
                stack.push(root);
                root = root.left;
            }else{
                TreeNode top = stack.peek();
                if(top.right != null && top.right != prev){
                    root = top.right;
                }else{
                    stack.pop();
                    res.add(top.val);
                    prev = top;
                }
            }
        }
        return res;
        
    }
    //拓展
    //recover inorder based on given preorder 
    //Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree.
    //You may assume each number in the sequence is unique.
    public boolean verifyPreorder(int[] preorder) {
        Stack stack = new Stack();
        Stack inorder = new Stack();
        for(int v:preorder){
            if(!inorder.isEmpty() && v < inorder.peek())
                return false;
            while(!stack.isEmpty() && stack.peek() < v){
                inorder.push(stack.pop());
            }
            stack.push(v);
        }
        return true;
    }