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?
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SOLUTION 1:
递归解法
public List<Integer> postorderTraversal1(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
dfs(root, ret);
return ret;
}
// Solution 1: rec
public void dfs(TreeNode root, List<Integer> ret) {
if (root == null) {
return;
}
dfs(root.left, ret);
dfs(root.right, ret);
ret.add(root.val);
}
SOLUTION 2:
/**
* 后序遍历迭代解法
* http://www.youtube.com/watch?v=hv-mJUs5mvU
* http://blog.csdn.net/tang_jin2015/article/details/8545457
* 从左到右的后序 与从右到左的前序的逆序是一样的，所以就简单喽！ 哈哈
* 用另外一个栈进行翻转即可喽
*/
// Solution 2: iterator
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
if (root == null) {
return ret;
} Stack<TreeNode> s = new Stack<TreeNode>();
Stack<Integer> out = new Stack<Integer>(); s.push(root); while (!s.isEmpty()) {
TreeNode cur = s.pop();
out.push(cur.val); if (cur.left != null) {
s.push(cur.left);
} if (cur.right != null) {
s.push(cur.right);
}
} while (!out.isEmpty()) {
ret.add(out.pop());
} return ret;
}
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/PostorderTraversal.java