题目:
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1
\
2
/
3
return [3,2,1]
.
Note: Recursive solution is trivial, could you do it iteratively?
题解:
递归方法代码:
1 public void helper(TreeNode root, ArrayList<Integer> re){
2 if(root==null)
3 return;
4
5 helper(root.left,re);
6 helper(root.right,re);
7 re.add(root.val);
8 }
9 public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> re = new ArrayList<Integer>();
if(root==null)
return re;
helper(root,re);
return re;
}
非递归方法代码:
引用自Code ganker:http://blog.csdn.net/linhuanmars/article/details/22009351
“接下来是迭代的做法,本质就是用一个栈来模拟递归的过程,但是相比于Binary
Tree Inorder Traversal和Binary
Tree Preorder Traversal,后序遍历的情况就复杂多了。我们需要维护当前遍历的cur指针和前一个遍历的pre指针来追溯当前的情况(注意这里是遍历的指针,并不是真正按后序访问顺序的结点)。具体分为几种情况:
(1)如果pre的左孩子或者右孩子是cur,那么说明遍历在往下走,按访问顺序继续,即如果有左孩子,则是左孩子进栈,否则如果有右孩子,则是右孩子进栈,如果左右孩子都没有,则说明该结点是叶子,可以直接访问并把结点出栈了。
(2)如果反过来,cur的左孩子是pre,则说明已经在回溯往上走了,但是我们知道后序遍历要左右孩子走完才可以访问自己,所以这里如果有右孩子还需要把右孩子进栈,否则说明已经到自己了,可以访问并且出栈了。
(3)如果cur的右孩子是pre,那么说明左右孩子都访问结束了,可以轮到自己了,访问并且出栈即可。
算法时间复杂度也是O(n),空间复杂度是栈的大小O(logn)。实现的代码如下(代码引用自:http://www.programcreek.com/2012/12/leetcode-solution-of-iterative-binary-tree-postorder-traversal-in-java/):”
1 public ArrayList<Integer> postorderTraversal(TreeNode root) {
2
3 ArrayList<Integer> lst = new ArrayList<Integer>();
4
5 if(root == null)
6 return lst;
7
8 Stack<TreeNode> stack = new Stack<TreeNode>();
9 stack.push(root);
TreeNode prev = null;
while(!stack.empty()){
TreeNode curr = stack.peek();
// go down the tree.
//check if current node is leaf, if so, process it and pop stack,
//otherwise, keep going down
if(prev == null || prev.left == curr || prev.right == curr){
//prev == null is the situation for the root node
if(curr.left != null){
stack.push(curr.left);
}else if(curr.right != null){
stack.push(curr.right);
}else{
stack.pop();
lst.add(curr.val);
}
//go up the tree from left node
//need to check if there is a right child
//if yes, push it to stack
//otherwise, process parent and pop stack
}else if(curr.left == prev){
if(curr.right != null){
stack.push(curr.right);
}else{
stack.pop();
lst.add(curr.val);
}
//go up the tree from right node
//after coming back from right node, process parent node and pop stack.
}else if(curr.right == prev){
stack.pop();
lst.add(curr.val);
}
prev = curr;
}
return lst;
}