leetcode -- Binary Tree Postorder Traversal
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?
[解题思路]
后序遍历的非递归相对来说比较难,根节点需要在其左右孩子都访问结束后才能被访问,因此对于任一节点,先将其入栈,如果p不存在左孩子和右孩子,则可以直接访问它;或者p存在左孩子或者右孩子,但是其左孩子和右孩子都已被访问过了,则同样可以直接访问该节点。若非上述两种情况,则将p的右孩子和左孩子依次入栈,这样就保证了每次去站定元素的时候,左孩子在右孩子前面被访问,左孩子和右孩子在根节点前面被访问。
1 public ArrayList<Integer> postorderTraversal(TreeNode root) { 2 // IMPORTANT: Please reset any member data you declared, as 3 // the same Solution instance will be reused for each test case. 4 ArrayList<Integer> result = new ArrayList<Integer>(); 5 if(root == null){ 6 return result; 7 } 8 9 Stack<TreeNode> stack = new Stack<TreeNode>(); 10 TreeNode cur = null, pre = null; 11 stack.push(root); 12 13 while(!stack.empty()){ 14 cur = stack.peek(); 15 if((cur.left == null && cur.right == null) || 16 ((pre != null) && (cur.left == pre || cur.right == pre))){ 17 result.add(cur.val); 18 pre = cur; 19 stack.pop(); 20 } else { 21 if(cur.right != null){ 22 stack.push(cur.right); 23 } 24 if(cur.left != null){ 25 stack.push(cur.left); 26 } 27 } 28 29 } 30 31 return result; 32 }