513.找树左下角的值
public int findBottomLeftValue(TreeNode root) {List<Integer> path = new ArrayList<>();List<List<Integer>> ans = new ArrayList<>();Boolean[] flag = new Boolean[1];flag[0] = true;dfs(root, path, ans, flag);int maxSizeIndex = 0;if (ans.size() == 0) {return -1;}int maxSize = ans.get(0).size();for (int i = 0; i < ans.size(); i++) {if (maxSize < ans.get(i).size()) {maxSize = ans.get(i).size();maxSizeIndex = i;}}if (ans.get(maxSizeIndex).size() == 0) {return -1;}return ans.get(maxSizeIndex).get(ans.get(maxSizeIndex).size() - 1);
}public void dfs(TreeNode root, List<Integer> path, List<List<Integer>> ans, Boolean[] flag) {if (root == null) {return;}path.add(root.val);if (root.left == null && root.right == null) {ans.add(new ArrayList<Integer>(path));}// flag[0] = true;dfs(root.left, path, ans, flag);// flag[0] = false;dfs(root.right, path, ans, flag);path.remove(path.size() - 1);}
112. 路径总和
public boolean hasPathSum(TreeNode root, int targetSum) {List<Integer> path = new ArrayList<>();List<List<Integer>> ans = new ArrayList<>();dfs(root, path, ans);for (List<Integer> an : ans) {int sum = 0;for (Integer integer : an) {sum += integer;}if (sum == targetSum) {return true;}}return false;
}public void dfs(TreeNode root, List<Integer> path, List<List<Integer>> ans) {if (root == null) {return;}path.add(root.val);if (root.left == null && root.right == null) {ans.add(new ArrayList<Integer>(path));}dfs(root.left, path, ans);dfs(root.right, path, ans);path.remove(path.size() - 1);}
105.从前序与中序遍历序列构造二叉树
public TreeNode buildTree(int[] preorder, int[] inorder) {Map<Integer, Integer> map = new HashMap<>();for (int i = 0; i < inorder.length; i++) {map.put(inorder[i], i);}return build(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1, map);
}public TreeNode build(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd, Map<Integer, Integer> inMap) {if (preStart > preEnd || inStart > inEnd) {return null;}int rootVal = preorder[preStart];TreeNode root = new TreeNode(rootVal);Integer rootIndex = inMap.get(rootVal);//左子树的长度int instance = rootIndex - inStart;//左递归,分解成子问题,每次递归去除根节点,保留左子树的节点 root.left = build(preorder, inorder, preStart + 1, preStart + instance, inStart, inStart + instance - 1, inMap);//右递归,分解成子问题,每次递归去除根节点,保留右子树的节点 root.right = build(preorder, inorder, preStart + instance + 1, preEnd, inStart + instance + 1, inEnd, inMap);return root;
}
)
