一篇文章教你如何用多种迭代写法实现二叉树遍历

目录

思想

利用栈和队列都可以实现树的迭代遍历。递归的写法将这个遍历的过程交给系统的堆栈去实现了，所以思想都是一样的、无非就是插入值的时机不一样。利用栈的先进先出的特点，对于前序遍历、我们可以先将当前的值放进结果集中，表示的是根节点的值、然后将当前的节点加入到栈中、当前的节点等于自己的left、再次循环的时候、也会将left作为新的节点、直到节点为空、也就是走到了树的最左边、然后回退、也就是弹栈、、也可以认为回退的过程是从低向上的、具体就是让当前的节点等于栈弹出的right、继续重复上面的过程，也就实现了树的前序遍历、也就是bfs.后续遍历、中序遍历思想也是类似的。

实现

    public list<integer> preordertraversal1(treenode root) {
        list<integer> res = new arraylist<>();
        stack<treenode> stack = new stack<>();
        while (!stack.isempty() || root != null) {
            while (root != null) {
                res.add(root.val);
                stack.add(root);
                root = root.left;
            }
            treenode cur = stack.pop();
            root = cur.right;
        }
        return res;
    }
    public list<integer> preordertraversal2(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.add(root);
                root = root.left;
            } else {
                treenode cur = stack.pop();
                root = cur.right;
            }
        }
        return res;
    }
    public list<integer> preordertraversal3(treenode root) {
        list<integer> res = new arraylist<>();
        if (root == null) return res;
        stack<treenode> stack = new stack<>();
        stack.push(root);
        while (!stack.isempty()) {
            treenode cur = stack.pop();
            res.add(cur.val);
            if (cur.right != null) {
                stack.push(cur.right);
            }
            if (cur.left != null) {
                stack.push(cur.left);
            }
        }
        return res;
    }
    public list<integer> preordertraversal4(treenode root) {
        list<integer> res = new arraylist<>();
        if (root == null) {
            return res;
        }
        linkedlist<treenode> queue = new linkedlist<>();
        queue.add(root);
        while (!queue.isempty()) {
            root = queue.poll();
            res.add(root.val);
            if (root.right != null) {
                queue.addfirst(root.right);
            }
            if (root.left != null) {
                root = root.left;
                while (root != null) {
                    res.add(root.val);
                    if (root.right != null) {
                        queue.addfirst(root.right);
                    }
                    root = root.left;
                }
            }
        }
        return res;
    }
    public list<integer> inordertraversal1(treenode root) {
        list<integer> res = new arraylist<>();
        stack<treenode> stack = new stack<>();
        while (root != null || !stack.isempty()) {
            if (root != null) {
                stack.add(root);
                root = root.left;
            } else {
                treenode cur = stack.pop();
                res.add(cur.val);
                root = cur.right;
            }
        }
        return res;
    }
    public list<integer> inordertraversal2(treenode root) {
        list<integer> res = new arraylist<>();
        stack<treenode> stack = new stack<>();
        while (root != null || !stack.isempty()) {
            while (root != null) {
                stack.add(root);
                root = root.left;
            }
            treenode cur = stack.pop();
            res.add(cur.val);
            root = cur.right;
        }
        return res;
    }
    public list<integer> postordertraversal1(treenode root) {
        list<integer> res = new arraylist<>();
        if (root == null) return res;
        stack<treenode> stack = new stack<>();
        stack.push(root);
        while (!stack.isempty()) {
            treenode cur = stack.pop();
            res.add(cur.val);
            if (cur.left != null) {
                stack.push(cur.left);
            }
            if (cur.right != null) {
                stack.push(cur.right);
            }
        }
        collections.reverse(res);
        return res;
    }
    public list<integer> postordertraversal2(treenode root) {
        list<integer> res = new arraylist<>();
        stack<treenode> stack = new stack<>();
        while (!stack.isempty()) {
            while (root != null) {
                res.add(root.val);
                stack.push(root);
                root = root.right;
            }
            treenode cur = stack.pop();
            root = cur.left;
        }
        collections.reverse(res);
        return res;
    }
    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()){
            int size = queue.size();
            list<integer> list = new arraylist<>();
            while(size!=0){
                treenode cur = queue.poll();
                list.add(cur.val);
                if(cur.left!=null){
                    queue.offer(cur.left);
                }
                if(cur.right!= null){
                    queue.offer(cur.right);
                }
                size --;
            }
            ret.add(list);
        }
        return ret;
    }

总结

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