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Invert a binary tree. 4 / \ 2 7 / \ / \1 3 6 9to 4 / \ 7 2 / \ / \9 6 3 1方法：先创建一个空结点，将左右结点交换，再用递归代码：/** * Definition for a binary

public class Solution {    public boolean isSameTree(TreeNode p, TreeNode q) {        if(p==null&&q==null){            return true;        }else if(p==null||q==null){            return false

public class Solution {    public int maxDepth(TreeNode root) {        if(root == null) return 0;        if(root.left==null&&root.right==null) return 1;        int x = Math.max(maxDepth(root.l

public class Solution {    public TreeNode sortedArrayToBST(int[] nums) {        return build(nums,0,nums.length-1);    }        public TreeNode build(int[]nums,int left,int right){       

Given a binary tree, return the postorder traversal of its nodes' values.For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3return [3,2,1].代码：/** * Defini

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Sol

Note:You may only use constant extra space.You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).For example,Given the

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.For example:Given the below binary tree and sum

1：由于要求链表是有序的，可以借助二叉树中序遍历，因为中序遍历算法的特点就是从小到大访问结点。当遍历访问到根结点时，假设根结点的左侧已经处理好，只需将根结点与上次访问的最近结点（左子树中最大值结点）的指针连接好即可。进而更新当前链表的最后一个结点指针。2：由于中序遍历过程正好是转换成链表的过程，即可采用递归处理struct BinaryTreeNode  {

For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3return [1,3,2].public class Solution {    public List inorderTraversal(TreeNode root) {        List res = new