LeetCode 之 preorder traversal of binary tree

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Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

两种解法，一种是递归一种是迭代

线上节点类：

package node;public class TreeNode {private int val;private TreeNode left;private TreeNode right;public TreeNode(int x) {val = x;}public int getVal(){return this.val;}public void setVal(int val){this.val = val;}public TreeNode getLeft(){return this.left;}public TreeNode getRight(){return this.right;}public void setLeft(TreeNode left){this.left = left;}public void setRight(TreeNode right){this.right = right;}}

解题：

（1）递归

    public ArrayList<Integer> preorderTraversal(TreeNode root) {ArrayList<Integer> result = new ArrayList<Integer>();if (root != null) {result.addAll(preorderTraversal(root.getLeft()));result.add(root.getVal());result.addAll(preorderTraversal(root.getRight()));}return result;    }

（2）迭代

迭代使用stack 来实现，先进后出。

 public ArrayList<Integer> preOrder(TreeNode root){    ArrayList<Integer> result = new ArrayList<Integer>();    Stack<TreeNode> nodeStack = new Stack<TreeNode>();      nodeStack.push(root);    while(!nodeStack.isEmpty()){    TreeNode node = nodeStack.pop();    result.add(node.getVal());        if(node.getRight()!=null){    nodeStack.push(node.getRight());    }        if(node.getLeft()!=null){    nodeStack.push(node.getLeft());    }    }        return result;    }

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