leetcodes树操作习题_private int〔〕path=-优快云博客

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leetcode树习题

104：二叉树的深度

递归并借用数学函数解决

public int maxDepth(TreeNode root) {
    if (root == null) return 0;
    return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
}

110:平衡树

private boolean result = true;

public boolean isBalanced(TreeNode root) {
    maxDepth(root);
    return result;
}

public int maxDepth(TreeNode root) {
    if (root == null) return 0;
    int l = maxDepth(root.left);
    int r = maxDepth(root.right);
    if (Math.abs(l - r) > 1) result = false;
    return 1 + Math.max(l, r);
}

543：二叉树的直径

理解直径的概念

private int max = 0;

public int diameterOfBinaryTree(TreeNode root) {
    depth(root);
    return max;
}

private int depth(TreeNode root) {
    if (root == null) return 0;
    int l = depth(root.left);
    int r = depth(root.right);
    max = Math.max(max, l + r);
    return Math.max(l, r) + 1;
}

226：翻转树

递归思想

 public TreeNode invertTree(TreeNode root) {
         if(root==null) return root;
         TreeNode left=root.left;        
         TreeNode right=root.right;                   
         root.left=invertTree(right);        
         root.right=invertTree(left);        
         return root;   
          }

617：归并两棵树

递归思想

public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
    if (t1 == null && t2 == null) return null;
    if (t1 == null) return t2;
    if (t2 == null) return t1;
    TreeNode root = new TreeNode(t1.val + t2.val);
    root.left = mergeTrees(t1.left, t2.left);
    root.right = mergeTrees(t1.right, t2.right);
    return root;
}

112：判断路径和是否等于一个数

递归

public boolean hasPathSum(TreeNode root, int sum) {
    if (root == null) return false;
    if (root.left == null && root.right == null && root.val == sum) return true;
    return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}

对于递归过程中是否需要记录变量值，不是很清楚，需要继续总结摸索。

437：统计路径满足某数的数量

如何完成遍历，双递归操作

public int pathSum(TreeNode root, int sum) {
    if (root == null) return 0;
    int ret = pathSumStartWithRoot(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    return ret;
}

private int pathSumStartWithRoot(TreeNode root, int sum) {
    if (root == null) return 0;
    int ret = 0;
    if (root.val == sum) ret++;
    ret += pathSumStartWithRoot(root.left, sum - root.val) + pathSumStartWithRoot(root.right, sum - root.val);
    return ret;
}

572：子树

和寻找路径数思路相同
注意题干中要求是非空子树

public boolean isSymmetric(TreeNode root) {
    if (root == null) return true;
    return isSymmetric(root.left, root.right);
}

private boolean isSymmetric(TreeNode t1, TreeNode t2) {
    if (t1 == null && t2 == null) return true;
    if (t1 == null || t2 == null) return false;
    if (t1.val != t2.val) return false;
    return isSymmetric(t1.left, t2.right) && isSymmetric(t1.right, t2.left);
}

111：最小路径问题

注意没有左右子树的情况

public int minDepth(TreeNode root) {
    if (root == null) return 0;
    int left = minDepth(root.left);
    int right = minDepth(root.right);
    if (left == 0 || right == 0) return left + right + 1;
    return Math.min(left, right) + 1;
}

404：统计左叶子节点的和

判断左侧是否为叶子节点，找到左侧的叶子节点为结束标志
叶子节点的判断是关键

public int sumOfLeftLeaves(TreeNode root) {
    if (root == null) return 0;
    if (isLeaf(root.left)) return root.left.val + sumOfLeftLeaves(root.right);
    return sumOfLeftLeaves(root.left) + sumOfLeftLeaves(root.right);
}

private boolean isLeaf(TreeNode node){
    if (node == null) return false;
    return node.left == null && node.right == null;
}

687：相同节点值的最大路径长度

递归还是解决问题的难点
区分开，记录值和输出贡献值

private int path = 0;

public int longestUnivaluePath(TreeNode root) {
    dfs(root);
    return path;
}

private int dfs(TreeNode root){
    if (root == null) return 0;
    int left = dfs(root.left);
    int right = dfs(root.right);
    int leftPath = root.left != null && root.left.val == root.val ? left + 1 : 0;
    int rightPath = root.right != null && root.right.val == root.val ? right + 1 : 0;
    path = Math.max(path, leftPath + rightPath);
    return Math.max(leftPath, rightPath);

337：间隔遍历

理解题意是关键，间隔一层才能取得最大的收益为什么呢？

public int rob(TreeNode root) {
    if (root == null) return 0;
    int val1 = root.val;
    if (root.left != null) val1 += rob(root.left.left) + rob(root.left.right);
    if (root.right != null) val1 += rob(root.right.left) + rob(root.right.right);
    int val2 = rob(root.left) + rob(root.right);
    return Math.max(val1, val2);
}